• Algebra > Cubic Equations

Cubic Equations

The solution of the quadratic equation is known. The famous quadratic formula gives us two roots of the quadratic equation of the form \( ax^{2} + bx + c = 0 \): \( x = \dfrac{- b \pm \sqrt{b^{2} - 4ac}}{2a} \).

To tackle the cubic equation of the form \( Ax^{3} + Bx^{2} + Cx + D = 0 \) is a challenge, however. There is a general cubic roots formula, but it is not as elegant as the quadratic formula and does not give the roots in the form we need. We usually look for real roots that do not contain nested complex numbers, i.e., we prefer complex roots in the form of \(a + bi \). The cubic formula does not always produce the result in the form we seek.

Cubics of the form \( Ax^{3} + Bx^{2} + Cx + D = 0 \) can be reduced or ‘depressed’ to the form \( {x^{3}} + Bx = C \). Therefore, we will focus on solving the cubic equation of this form here. Even the solution to this cubic does not always give us the solution in the form we require. However, we can, at the least, get one cubic root of the depressed cubic polynomial. Once we have one root, finding the other 2 roots is much easier.

Note that we let the coefficient of \( x^{3} \) term equal 1. If it is not, divide the whole equation by the coefficient make it simpler.

Solution to \( {x^{3}} + Bx = C \)

Let the solution of this form of the cubic be \( x = m^{1/3} - n^{1/3} \). Then

(i) \( x^{3} = m - 3m^{2/3}n^{1/3} + 3m^{1/3}n^{2/3} - n \)

(ii) \( x^{3} = (m - n) - 3(mn)^{1/3}(m^{1/3} - n^{1/3}) \)

(iii) \( x^{3} + 3(mn)^{1/3}x = m - n \)

The equation derived in (iii) is satisfied when \( B = 3(mn)^{1/3} \) and \( m - n = C \). Solving for mn, the following is obtained:

(iv) \( mn = \frac{B^{3}}{27} \)

Now, \( C^{2} = m^{2} - 2mn + n^{2} \). Adding 4mn to both sides produces the equation:

(v) \( C^{2} + \frac{4B^{3}}{27} = m^{2} + 2mn + n^{2} \)

(vi) \( C^{2} + \frac{4B^{3}}{27} = (m + n)^{2} \)

(vii) \( m + n = \pm \sqrt{\frac{4B^{3}}{27} + C^{2}} \)

But, \( m - n = C \). Solving for m and n using the system of equations:

(viii) \( m = \frac{1}{2}\left( C \pm \sqrt{\frac{4B^{3}}{27} + C^{2}} \right) \)

(ix) \( n = \frac{1}{2}\left( - C \pm \sqrt{\frac{4B^{3}}{27} + C^{2}} \right) \)

As we stated in the beginning that the solution is of the form \( x = m^{1/3} - n^{1/3} \), we can use the values of m and n obtained in terms of B and C to yield the solution:

The solution of the cubic equation of the form \( x^{3} + Bx = C \) is \( x = \sqrt[3]{\frac{1}{2}C \pm \frac{1}{18}\sqrt{12B^{3} + 81C^{2}}} \, - \) \( \sqrt[3]{ - \frac{1}{2}C \pm \frac{1}{18}\sqrt{12B^{3} + 81C^{2}}} \).

Limitations: If \( 12B^{3} + 81C^{2} \) is not negative, we get a solution that has nested square roots. If not, we get a solution that is a combination of the cube root of complex numbers, which itself is difficult to resolve but not impossible in some cases. Regardless, this one root, I believe, represents the one real root of the cubic equation, since all cubic equations have one real root. The other two real or non-real roots can be determined easily after this.

Which Sign?

The cubic solution gives only one solution. However, we have a plus/minus sign inside the two cube roots in the solution. We have to pick one. How do we pick?

It seems to be best to choose the sign that maintains a positive number inside the cube root. For example \( \sqrt[3]{-8} \) can be written as \( -\sqrt[3]{8} \), so that the positive number is inside the cube root. When checking your answer with Wolfram, this works best.

Also remember the solution is of the form \( m^{1/3} - n^{1/3} \), where m and n are the radicands and \( m - n = C \). Keep the sign the same for the cube root of both radicands initially. Once picked, you can take the negative of radicands to bring the sign out so that the solution is in the form of \( m^{1/3} + n^{1/3} \) and \( m - n \) no longer equals C. In this case, \( m - n \) will likely equal 0 or \( m + n = C \).

Example 1 will be a straight-forward example. Example 2 will require some sign manipulation.

EXAMPLES

Example 1

Find the real root of \( x^{3} + 4x = 8 \).

(i) \( x = \sqrt[3]{\frac{1}{2}C \pm \frac{1}{18}\sqrt{12B^{3} + 81C^{2}}} \, - \) \(\sqrt[3]{ - \frac{1}{2}C \pm \frac{1}{18}\sqrt{12B^{3} + 81C^{2}}} \)

(ii) \( x = \sqrt[3]{\frac{1}{2}\cdot 8 \pm \frac{1}{18}\sqrt{12(4)^{3} + 81(8)^{2}}} \, - \) \( \sqrt[3]{ - \frac{1}{2}\cdot 8 \pm \frac{1}{18}\sqrt{12(4)^{3} + 81(8)^{2}}} \)

(iii) \( x = \sqrt[3]{4 \pm \frac{1}{18}\sqrt{12\cdot 64 + 81\cdot 64}} \, - \) \( \sqrt[3]{ -4 \pm \frac{1}{18}\sqrt{12\cdot 64 + 81\cdot 64}} \)

(iv) \( x = \sqrt[3]{4 \pm \frac{4}{9}\sqrt{93}} \, - \) \( \sqrt[3]{ -4 \pm \frac{4}{9}\sqrt{93}} \)

Let’s choose the positive sign, giving us \( x = \sqrt[3]{4 + \frac{4}{9}\sqrt{93}} \, - \) \( \sqrt[3]{ -4 + \frac{4}{9}\sqrt{93}} \approx 1.3647 \). Both \( 4 + \frac{4}{9}\sqrt{93} \) and \( -4 + \frac{4}{9}\sqrt{93} \) are positive. Therefore, this answer is correct. This answer is confirmed with Wolfram Alpha.

If we had chose the negative sign, we would get \( x = \sqrt[3]{4 - \frac{4}{9}\sqrt{93}} \, - \) \( \sqrt[3]{ -4 - \frac{4}{9}\sqrt{93}} \). Both expressions in the cubic are negative. If we change them to be positive inside, we get our original solution. Also notice that \(m - n = 4 + \frac{4}{9}\sqrt{93} - (-4 + \frac{4}{9}\sqrt{93}) = 8 \).

This cubic has 1 real solution and 2 non-real solutions. This can be seen by graphing it.

Example 1 was straight-forward. We got the answer, chose the sign and it was the root. Example 2 requires a bit of sign manipulation.

Example 2

Find the roots of \( x^{3} - 4x = 8 \). This cubic has 1 real root when graphed. If we apply the formula, we get:

(i) \( x = \sqrt[3]{\frac{1}{2}\cdot 8 \pm \frac{1}{18}\sqrt{12\cdot(-4)^{3} + 81(8)^{2}}} \, - \) \(\sqrt[3]{ - \frac{1}{2}\cdot 8 \pm \frac{1}{18}\sqrt{12(-4)^{3} + 81(8)^{2}}} \)

(ii) \( x = \sqrt[3]{4 \pm \frac{1}{18}\sqrt{64(81-12)}} \, - \) \(\sqrt[3]{ -4 \pm \frac{1}{18}\sqrt{64(81-12)}} \)

(iii) \( x = \sqrt[3]{4 \pm \frac{4}{9}\sqrt{69}} \, - \) \(\sqrt[3]{ -4 \pm \frac{4}{9}\sqrt{69}} \)

Let’s try the positive sign first, which guarantees the first radicand is positive. Taking the positive sign, the root is \( x = \sqrt[3]{4 + \frac{4}{9}\sqrt{69}} \, - \) \(\sqrt[3]{ -4 + \frac{4}{9}\sqrt{69}} \). However, \( \sqrt[3]{-4+\frac{4}{9}\sqrt{69}} \) has a negative number inside the cube root. Unlike negative square roots, negative cube roots can have real values which are negative. Hence, \( \sqrt[3]{-4+\frac{4}{9}\sqrt{69}} \) is the same as \( -\sqrt[3]{4-\frac{4}{9}\sqrt{69}} \). Therefore, the real solution to this cubic equation is \( x = \sqrt[3]{4+\frac{4}{9}\sqrt{69}} + \sqrt[3]{4-\frac{4}{9}\sqrt{69}} \) which is approximately 2.6494. This solution checks.

Since we changed the sign, in this case \( m + n = 8 \).

For some reason, Wolfram gives the cube root of \( \sqrt[3]{-4+\frac{4}{9}\sqrt{69}} \) as a complex number instead of the negative real value. That must be kept in consideration when checking your answer. This is why I have tried to keep the radicands positive.

THE OTHER ROOTS

We found a real root of \( x^{3} + 4x = 8 \). What about the other real or non-real roots? Assuming there is only 1 real root, let’s find the non-real roots using the equation \( x^{3} + Bx = C \). Suppose the real root is \( r_{1} \) from the formula above.

We know that the other roots must be of the form \( a + bi \) and \( a - bi \).

We know that the sum of the roots is 0 because the \( x^{2} \) term is missing. Hence, \( r_{1} + a + bi + a - bi = 0 \). Or \( a = -\frac{1}{2}r_{1} \). That was easy - we already found a in our non-real roots without much effort. Now, we need to find b.

We also know that the product of the roots is C. So \( r_{1}(a - bi)(a+bi) = C \). (To review how sum of the roots and product of the roots are related to the coefficients of the polynomial, go to Roots of a Polynomial.) Let’s work with this a bit.

(i) \( r_{1}(a^{2} + b^{2}) = C \)

(ii) \( r_{1}(\frac{1}{4}r_{1}^{2} + b^{2}) = C \)  (Substituting \( a = -\frac{1}{2}r_{1} \))

(iii) \( \frac{1}{4}r_{1}^{3} + r_{1}b^{2} = C \)

(iv) \( r_{1}b^{2} = C - \frac{1}{4}r_{1}^{3} \)

(v) \( b^{2} = \frac{C - \frac{1}{4}r_{1}^{3}}{r_{1}} \)

(vi) \( b = \pm\sqrt{\frac{C - \frac{1}{4}r_{1}^{3}}{r_{1}}} \)

(vii) \( b = \pm\frac{1}{2}\sqrt{\frac{4C - r_{1}^{3}}{r_{1}}} \)

Since \( r_{1} \) is a root of the cubic equation, this means that \( r_{1}^{3} = -Br_{1} + C \). Therefore, (vii) is the same as:

(viii) \( b = \pm\frac{1}{2}\sqrt{\frac{4C - (-Br_{1} + C)}{r_{1}}} \)

(ix) \( b = \pm\frac{1}{2}\sqrt{\frac{3C + Br_{1}}{r_{1}}} \)

If we put together the solution for a and b in (vii), we get: \( -\frac{1}{2}r_{1} \pm\frac{1}{2}\sqrt{\frac{4C - r_{1}^{3}}{r_{1}}}i \), which can also be written as \( -\frac{1}{2}r_{1} \pm\frac{1}{2}\sqrt{\frac{-4C + r_{1}^{3}}{r_{1}}} \) if we want to avoid imaginary numbers.

If we put together the solution for a and b in (ix), we get \( -\frac{1}{2}r_{1} \pm\frac{1}{2}\sqrt{\frac{3C + Br_{1}}{r_{1}}}i \).

Both of these solutions are useful depending on whether the other 2 solutions are real or non-real. If the other two solutions are real, then \( -\frac{1}{2}r_{1} \pm\frac{1}{2}\sqrt{\frac{-4C + r_{1}^{3}}{r_{1}}} \) is helpful. If the solutions are non-real, then \( -\frac{1}{2}r_{1} \pm\frac{1}{2}\sqrt{\frac{3C + Br_{1}}{r_{1}}}i \) is helpful because it gives us the solutions in the form \( a + bi \) rather than nested complex numbers.

We do not actually need to know if the cubic equation has 3 real roots or 1 real and 2 non-real roots. We can use either formula. With complex number calculations, the result we arrive at should be the appropriate solution, as in Example 3.

Example 3

Given the real root of \( x^{3} - 15x - 4 = 0 \) is 4, find the other real or non-real roots.

This cubic does have 3 real roots, with 4 being the obvious one that can be found by trial and error. Let’s use the formula to find the other roots and see if they turn out to be real or non-real.

(i) \( -\frac{1}{2}r_{1} \pm\frac{1}{2}\sqrt{\frac{3C + Br_{1}}{r_{1}}}i \)

(i) \(x = -\frac{1}{2}(4) \pm \frac{1}{2}\sqrt{\frac{3(4) + (-15)(4)}{4}}i \)

(ii) \(x = -2 \pm \frac{1}{2}(\sqrt{-12})i \)

(iii) \(x = -2 \pm \frac{1}{2}(2\sqrt{3}i)i \)

(iv) \(x = -2 \pm \sqrt{3}i^{2} \)

(v) \(x = -2 \pm \sqrt{3} \)

Hence, the roots of \( x^{3} - 15x - 4 = 0 \) are 4, \( -2 + \sqrt{3} \), and \( -2 - \sqrt{3} \).

The sum of the roots is 0, and the product is \( 4(-2+\sqrt{3})(-2-\sqrt{3}) \, = \) \( 4(4-3) = 4 \). Even though we used the formula with a complex number, all roots turned out to be real because the complex numbers multiplied to be -1. If we used the formula \( -\frac{1}{2}r_{1} \pm\frac{1}{2}\sqrt{\frac{-4C + r_{1}^{3}}{r_{1}}} \), we get the same answer and avoid calculations with imaginary numbers.

Figure 3 shows the 3 roots of \( f(x) = x^3 - 15x - 4 \). The function has a very high amplitude, so only a portion of the graph where the roots are is shown. The roots shown in decimal form are what we obtained.

What if we use \( -2 + \sqrt{3} \) in the formula? Let’s try that.

(i) \( -\frac{1}{2}r_{1} \pm\frac{1}{2}\sqrt{\frac{3C + Br_{1}}{r_{1}}}i \)

(ii) \( -\frac{1}{2}(-2 + \sqrt{3}) \pm\frac{1}{2}\sqrt{\frac{3(4) + (-15)(-2 + \sqrt{3})}{-2 + \sqrt{3}}}i \)

(iii) \( 1 - \frac{\sqrt{3}}{2} \pm\frac{1}{2}\sqrt{\frac{42 - 15\sqrt{3}}{-2 + \sqrt{3}}}i \)

(iv) \( 1 - \frac{\sqrt{3}}{2} \pm\frac{1}{2}\sqrt{\frac{(42 - 15\sqrt{3})(-2-\sqrt{3})}{(-2 + \sqrt{3})(-2 - \sqrt{3})}}i \)

(v) \( 1 - \frac{\sqrt{3}}{2} \pm\frac{1}{2}\sqrt{\frac{-39 - 12\sqrt{3}}{1}}i \)

(vi) \( 1 - \frac{\sqrt{3}}{2} \pm\frac{1}{2}\sqrt{39 + 12\sqrt{3}} \)

Believe it or not, (vi) evaluates to \( -2 - \sqrt{3} \) and 4! Because \( \sqrt{39 + 12\sqrt{3}}\) is equal to \( 6 + \sqrt{3} \).

Example 4

Find the non-real roots of \( x^{3} - 4x = 8 \).

We already know from the graph it only has 1 real root, which we found above: \( \sqrt[3]{4+\frac{4}{9}\sqrt{69}} + \sqrt[3]{4-\frac{4}{9}\sqrt{69}} \). The other 2 are non-real. Plugging in this root into our formula, we get:

\( x = -\frac{1}{2}\left( \sqrt[3]{4+\frac{4}{9}\sqrt{69}} + \sqrt[3]{4-\frac{4}{9}\sqrt{69}} \right) \, \pm \) \( \frac{1}{2}\sqrt{\frac{24-4\left(\sqrt[3]{4+\frac{4}{9}\sqrt{69}} + \sqrt[3]{4-\frac{4}{9}\sqrt{69}}\right)}{\sqrt[3]{4+\frac{4}{9}\sqrt{69}} + \sqrt[3]{4-\frac{4}{9}\sqrt{69}}}}i\)

What a monster solution! But the beauty is that this is completely in the form \( a + bi \) without any complex numbers nested in the roots. It evaluates to approximately -1.32471 ± 1.12456i.

Unlike Example 3 where the square root evaluated to an imaginary number and hence the root turned out to be real; in this example, everything inside the square root is a real number, allowing the root to be expressed as \( a + bi \). Therefore, the solution expressed as \( -\frac{1}{2}r_{1} \pm\frac{1}{2}\sqrt{\frac{3C + Br_{1}}{r_{1}}}i \) proves more useful in this case when we want to express our root as \( a + bi \).

CASUS IRRREDUCIBILIS

If we try the cubic formula on the cubic \( x^{3} - 4x = -2 \), which has 3 real roots, the solution gives us a non-real answer which is seemingly not a root.

(i) \( x = \sqrt[3]{\frac{1}{2}(-2) \pm \frac{1}{18}\sqrt{12(-4)^{3} + 81(-2)^{2}}} \, - \) \( \sqrt[3]{ - \frac{1}{2}(-2) \pm \frac{1}{18}\sqrt{12(-4)^{3} + 81(-2)^{2}}} \)

(ii) \( x = \sqrt[3]{-1 \pm \frac{1}{18}\sqrt{-444}} \, - \) \(\sqrt[3]{ 1 \pm \frac{1}{18}\sqrt{-444}} \)

(iii) \( x = \sqrt[3]{-1 \pm \frac{1}{9}\sqrt{-111}} \, - \) \(\sqrt[3]{ 1 \pm \frac{1}{9}\sqrt{-111}} \)

(iv) \( x = \sqrt[3]{-1 \pm \frac{\sqrt{111}}{9}i} \, - \) \(\sqrt[3]{ 1 \pm \frac{\sqrt{111}}{9}i} \)

Equation (iv) is indeed a solution because we can manipulate the solution of the form \( \sqrt[3]{a + bi} + \sqrt[3]{a - bi} \) so it reduces to the form \(s + ti + s - ti \), leaving \( 2s \) and the imaginary part cancels out. Therefore, we can change the signs so that m and n are conjugates and cube roots are added together rather than subtracted. Hence, when the cube roots of m and n are added, the imaginary part will cancel out.

If we choose the negative sign and write \( \sqrt[3]{-1 - \frac{\sqrt{111}}{9}i} \) as \( -\sqrt[3]{1 + \frac{\sqrt{111}}{9}i} \). Therefore, the solution is \( x = -\sqrt[3]{1 + \frac{\sqrt{111}}{9}i} \, - \) \(\sqrt[3]{ 1 - \frac{\sqrt{111}}{9}i} \, = \) \( x = -\left(\sqrt[3]{1 + \frac{\sqrt{111}}{9}i} + \sqrt[3]{ 1 - \frac{\sqrt{111}}{9}i}\right) \), which is approximately equal to -2.214. The complex expression turns into a real number.

When using the trigonometric identities, the solution is equal to \( -\frac{4\sqrt{3}}{3}\cos\left(\frac{1}{3}\tan^{-1}\frac{\sqrt{111}}{9}\right) \).

The real roots of this cubic are actually -2.2143..., 0.5392..., and 1.6751..., which are shown in the image above.

Although we can use trigonometry to reduce the complex expression to a real number, the fact that we arrived at nested complex numbers is referred to as casus irreducibilis. Sometimes if the nested complex number is simple, we can find the exact value. However, in this case, we can only approximate the complex value and put it in \( a + bi \) form using advanced CAS.

Remember the cubic \( x^{3} - 15x - 4 = 0 \). In the Notes section on the Wiki page of Casus Irreducibilis, a method is presented that can be used to find the roots. I must say that method is more complicated than the simple formula \( -\frac{1}{2}r_{1} \pm\frac{1}{2}\sqrt{\frac{4C - r_{1}^{3}}{r_{1}}}i \).

TRIGONOMETRIC SOLUTION

When we get casus irreducibilis, we can convert the result to a real number using trigonometry. A complex number to a fractional power can be represented as \( (a + bi)^{1/n} = (\sqrt{a^2+b^2})^{1/n}\cdot\left [\cos\left(\dfrac{\theta+2k\pi}{n}\right) + i\cdot\sin\left(\dfrac{\theta+2k\pi}{n}\right) \right ] \), where k is 0, 1, 2, ... n. When we have a root like \( x = -\sqrt[3]{1 + \frac{\sqrt{111}}{9}i} \, - \) \(\sqrt[3]{ 1 - \frac{\sqrt{111}}{9}i} \), it is of the form \( \sqrt[3]{a+bi} + \sqrt[3]{a-bi} \). The trigonometric conversion is:

(i) \( (\sqrt{a^{2}+b^{2}})^{1/n}\left[ \cos \left( \frac{1}{n}\tan^{-1}(\frac{b}{a})+\frac{2k\pi}{n} \right) + i\sin \left( \frac{1}{n}\tan^{-1}(\frac{b}{a}) +\frac{2k\pi}{n} \right) \right] \, + \) \( (\sqrt{a^{2}+b^{2}})^{1/n}\left[ \cos \left( \frac{1}{n}\tan^{-1}(-\frac{b}{a}) +\frac{2k\pi}{n} \right) + i\sin \left( \frac{1}{n}\tan^{-1}(-\frac{b}{a}) +\frac{2k\pi}{n} \right) \right] \)

Let’s use the sine and cosine sum identities to reduce this.

(ii) \( (\sqrt{a^{2}+b^{2}})^{1/n} \cos ( \frac{1}{n}\tan^{-1}(\frac{b}{a}))\cdot\cos(\frac{2k\pi}{n}) - \, \) \((\sqrt{a^{2}+b^{2}})^{1/n}\sin ( \frac{1}{n}\tan^{-1}(\frac{b}{a})) \cdot \sin(\frac{2k\pi}{n}) + \, \) \( (\sqrt{a^{2}+b^{2}})^{1/n}i\sin ( \frac{1}{n}\tan^{-1}(\frac{b}{a})) \cdot \cos(\frac{2k\pi}{n}) + \, \) \( (\sqrt{a^{2}+b^{2}})^{1/n}i \cos(\frac{1}{n}\tan^{-1}(\frac{b}{a})) \cdot \sin(\frac{2k\pi}{n}) \, + \) \( (\sqrt{a^{2}+b^{2}})^{1/n} \cos ( \frac{1}{n}\tan^{-1}(-\frac{b}{a}))\cdot\cos(\frac{2k\pi}{n}) - \, \) \( (\sqrt{a^{2}+b^{2}})^{1/n} \sin ( \frac{1}{n} \tan^{-1}(-\frac{b}{a})) \cdot \sin(\frac{2k\pi}{n}) + \, \) \( (\sqrt{a^{2}+b^{2}})^{1/n}i\sin ( \frac{1}{n}\tan^{-1}(-\frac{b}{a})) \cdot \cos(\frac{2k\pi}{n}) + \, \) \( (\sqrt{a^{2}+b^{2}})^{1/n}i \cos(\frac{1}{n}\tan^{-1}(-\frac{b}{a})) \cdot \sin(\frac{2k\pi}{n}) \)

(iii) \( (\sqrt{a^{2}+b^{2}})^{1/n} \cos ( \frac{1}{n}\tan^{-1}(\frac{b}{a}))\cdot\cos(\frac{2k\pi}{n}) - \, \) \((\sqrt{a^{2}+b^{2}})^{1/n}\sin ( \frac{1}{n}\tan^{-1}(\frac{b}{a})) \cdot \sin(\frac{2k\pi}{n}) + \, \) \( (\sqrt{a^{2}+b^{2}})^{1/n}i\sin ( \frac{1}{n}\tan^{-1}(\frac{b}{a})) \cdot \cos(\frac{2k\pi}{n}) + \, \) \( (\sqrt{a^{2}+b^{2}})^{1/n}i \cos(\frac{1}{n}\tan^{-1}(\frac{b}{a})) \cdot \sin(\frac{2k\pi}{n}) \, + \) \( (\sqrt{a^{2}+b^{2}})^{1/n} \cos ( \frac{1}{n}\tan^{-1}(\frac{b}{a}))\cdot\cos(\frac{2k\pi}{n}) + \, \) \( (\sqrt{a^{2}+b^{2}})^{1/n} \sin(\frac{1}{n} \tan^{-1}(\frac{b}{a})) \cdot \sin(\frac{2k\pi}{n}) - \, \) \( (\sqrt{a^{2}+b^{2}})^{1/n}i \sin( \frac{1}{n}\tan^{-1}(\frac{b}{a})) \cdot \cos(\frac{2k\pi}{n}) + \, \) \( (\sqrt{a^{2}+b^{2}})^{1/n}i \cos(\frac{1}{n}\tan^{-1}(\frac{b}{a})) \cdot \sin(\frac{2k\pi}{n}) \)

(iv) \( 2(\sqrt{a^{2}+b^{2}})^{1/n} \cos ( \frac{1}{n}\tan^{-1}(\frac{b}{a}))\cdot\cos(\frac{2k\pi}{n}) + \, \) \( 2(\sqrt{a^{2}+b^{2}})^{1/n}i \cos(\frac{1}{n}\tan^{-1}(\frac{b}{a})) \cdot \sin(\frac{2k\pi}{n}) \)

For our case, \( n = 3 \) for a cubic equation. And when \( k = 0 \), we have the following left that represents the real root of the cubic equation:

(v) \( 2(\sqrt{a^{2}+b^{2}})^{1/3}\left[ \cos \left( \frac{1}{3}\tan^{-1}(\frac{b}{a}) \right)\right] \)

Don’t confuse \( (\sqrt{a^{2}+b^{2}})^{1/3} \) as \( \sqrt[3]{a^{2}+b^{2}} \) because the former is equal to \( (a^{2}+b^{2})^{1/6} \) and the latter is equal to \( (a^{2}+b^{2})^{1/3} \).

Consider the cubic we encountered earlier \( x^{3} - 15x = 4 \). The solution we arrive at is \( x = \sqrt[3]{2 + 11i} + \sqrt[3]{2 - 11i} \), which is equal to 4. One of the cube roots of \( 2 + 11i \) is \( 2 + i \), and one of the cube roots of \( 2 - 11i \) is \( 2 - i \). Adding these together cancels out the imaginary part and leaves us with 4.

If we did not know it equals 4, we can represent it in trigonometric form as \( x = 2\sqrt{5}\cos(\frac{1}{3}\tan^{-1}\frac{11}{2}) = 4 \).

Estimation with calculator will give the answer as 4 or close to 4.

Because the real root is 4, \( \cos(\frac{1}{3}\tan^{-1}\frac{11}{2}) = \frac{2}{\sqrt{5}}\) and \( \sin(\frac{1}{3}\tan^{-1}\frac{11}{2}) = \frac{1}{\sqrt{5}} \), using the Pythagorean identity.

Let’s eliminate a and b and express the roots in terms of the coefficients. First, \( a = \frac{C}{2} \) and \( b = \frac{\sqrt{12B^{3}+81C^{2}}}{18} \). Therefore, \( a^{2} = \frac{C^{2}}{4} \) and \( b^{2} = \frac{12B^{3}+81C^{2}}{18^{2}} \).

If the cubic equation has 3 real roots, B cannot be positive. Therefore, \( \frac{\sqrt{12B^{3}+81C^{2}}}{18} \) will be equal to \( \frac{\sqrt{|12B^{3}-81C^{2}|}}{18} \) and \( b^{2} = \frac{|12B^{3}-81C^{2}|}{18^{2}} \).

Adding \( a^{2} \) and \( b^{2} \) gives us:

(i) \( \frac{C^{2}}{4} + \frac{12B^{3}-81C^{2}}{18^{2}} \)

(ii) \( \frac{18^{2}C^{2}}{4\cdot 18^{2}} + \frac{4\cdot 12B^{3}-4\cdot 81C^{2}}{4\cdot 18^{2}} \)

(iii) \( \frac{324C^{2} + 48B^{3}-324C^{2}}{4\cdot 18^{2}} \)

(iv) \( \frac{B^{3}}{27} \)

Therefore, \( 2(\sqrt{a^{2}+b^{2}})^{1/3} \) is equal to \( 2\sqrt{\frac{|B|}{3}} \).

Note: Our solution is in the form of \( \sqrt[3]{m} - \sqrt[3]{n} \). So we have to manipulate the signs so it is in the form of \( \sqrt[3]{m} + \sqrt[3]{n} \). However, the values of a and b should not change.

Notes

The above only works if a and b are not components of a complex number. But remember when we got three solutions to the trigonometric solution: one real and two non-real with k being 0, 1 and 2:

(iv) \( 2(\sqrt{a^{2}+b^{2}})^{1/3} \cos ( \frac{1}{3} \tan^{-1}(\frac{b}{a})) \cdot \cos(\frac{2k\pi}{3}) + \, \) \( 2(\sqrt{a^{2}+b^{2}})^{1/3}i \cos(\frac{1}{3} \tan^{-1}(\frac{b}{a})) \cdot \sin(\frac{2k\pi}{3}) \)

Letting k equal 0 gave us the real solution, which we discussed above. When k equals 1 and 2, we should get the 2 non-real solutions since the imaginary component does not equal 0.

When trying on \( x^{3} -4x = 8 \), which has 1 real root and 2 non-real roots with the real root being \( \sqrt[3]{4+\frac{4}{9}\sqrt{69}} + \sqrt[3]{4-\frac{4}{9}\sqrt{69}} \), if we let \( a = 4 \) and \( b = \frac{4}{9}\sqrt{69} \), we get a complex number and arctan function turns into arctanh function: \( 2\sqrt{\frac{4}{3}}\left[ \cos \left(\frac{1}{3}\tan^{-1}\frac{\sqrt{69}}{9}i\right) \right]\cos(\frac{2\pi}{3}) \approx -1.3247 \). This does give us the real part of the solution to the second root.

Evaluating the imaginary part gives us \( 2\sqrt{\frac{4}{3}}\left[ \cos \left(\frac{1}{3}\tan^{-1}\frac{\sqrt{69}}{9}i\right) \right]\sin(\frac{2\pi}{3}) \approx 2.2944 \), which is not correct. Oddly, \( 2\sqrt{\frac{4}{3}}\left[ \sin \left(\frac{1}{3}\tan^{-1}\frac{\sqrt{69}}{9}i\right) \right]\sin(\frac{2\pi}{3}) \approx 1.12456 \) is correct. I have not figured this out.

Example

Find the solution to \( x^{3} - 3x = -1 \).

Plugging into the formula, we get:

(i) \( \sqrt[3]{-\frac{1}{2} \pm \frac{\sqrt{3}}{2}i} - \sqrt[3]{\frac{1}{2} \pm \frac{\sqrt{3}}{2}i} \)

Choosing the sign is the challenge here. There does not seem to a clear method here. However, the actual solution after sign manipulation is: \( -\sqrt[3]{\frac{1}{2} - \frac{\sqrt{3}}{2}i} - \sqrt[3]{\frac{1}{2} + \frac{\sqrt{3}}{2}i} \, = \) \( -\left(\sqrt[3]{\frac{1}{2} - \frac{\sqrt{3}}{2}i} + \sqrt[3]{\frac{1}{2} + \frac{\sqrt{3}}{2}i}\right) \).

This is a case of casus irreducibilis. Hence, we can convert this to trigonometric form as \( -2\cos(\frac{1}{3}\tan^{-1}\sqrt{3}) \approx -1.8794 \).

The arctangent of \( \sqrt{3} \) is actually a known angle. It is equal to \( \frac{\pi}{3} \). Hence, the simplified solution is \( -2\cos\frac{\pi}{9} \).

Since this cubic equation has 3 real roots, the other solutions are \( -2\cos\frac{7\pi}{9} \), and \( -2\cos\frac{13\pi}{9} \).

THE FUTILITY OF THE THIRD-ANGLE IDENTITIES

As we mentioned earlier, to represent the solutions in algebraic form, we need to be able to find the third-angle value of sine and cosine. In this case, \( \cos\frac{\pi}{9} \) has an angle that is a third of \( \frac{\pi}{3} \). We know \( \cos\frac{\pi}{3} = \frac{1}{2} \), but there is no way to represent \( \cos\frac{\pi}{9} \) in algebraic form.

To find the algebraic form of \( \cos\frac{\pi}{9} \) requires us to solve the cubic equation \( 4x^{3} - 3x = \frac{1}{2} \), which is a solution to the triple-angle identity \( \cos\theta = 4\cos^{3}\frac{\theta}{3} - 3\cos\frac{\theta}{3} \).

Let’s divide by 4 and use our cubic roots formula on \( x^{3} - \frac{3}{4}x = \frac{1}{8} \):

(i) \( x = \sqrt[3]{\frac{1}{2}C \pm \frac{1}{18}\sqrt{12B^{3} + 81C^{2}}} \, - \) \( \sqrt[3]{ - \frac{1}{2}C \pm \frac{1}{18}\sqrt{12B^{3} + 81C^{2}}} \)

(ii) \( x = \sqrt[3]{\frac{1}{2}\cdot\frac{1}{8} \pm \frac{1}{18}\sqrt{12(-\frac{3}{4})^{3} + 81(\frac{1}{8})^{2}}} \, - \) \( \sqrt[3]{ - \frac{1}{2}\cdot \frac{1}{8} \pm \frac{1}{18}\sqrt{12(-\frac{3}{4})^{3} + 81(\frac{1}{8})^{2}}} \)

(iii) \( x = \sqrt[3]{\frac{1}{16} \pm \frac{1}{18}\sqrt{-12(\frac{27}{64}) + 81(\frac{1}{64})}} \, - \) \( \sqrt[3]{ - \frac{1}{16} \pm \frac{1}{18}\sqrt{-12(\frac{27}{64}) + 81(\frac{1}{64})}} \)

(iv) \( x = \sqrt[3]{\frac{1}{16} \pm \frac{1}{144}\sqrt{-243}} \, - \) \( \sqrt[3]{ - \frac{1}{16} \pm \frac{1}{144}\sqrt{-243}} \)

(v) \( x = \sqrt[3]{\frac{1}{16} \pm \frac{9}{144}\sqrt{-3}} - \sqrt[3]{ - \frac{1}{16} \pm \frac{9}{144}\sqrt{-3}} \)

(vi) \( x = \sqrt[3]{\frac{1}{16} \pm \frac{\sqrt{3}}{16}i} - \sqrt[3]{ - \frac{1}{16} \pm \frac{\sqrt{3}}{16}i} \)

(vii) \( x = \frac{1}{4}\sqrt[3]{1 \pm \sqrt{3}i} - \frac{1}{4}\sqrt[3]{ -1 \pm \sqrt{3}i} \)

This final expression we get in (vii) is not solvable since we cannot find the cube root of \( 1 + \sqrt{3}i \). The trigonometric equivalent of \( \sqrt[3]{1 \pm \sqrt{3}i} \) is \( \sqrt[3]{2}[\cos(\frac{1}{3}\tan^{-1}\sqrt{3}) \, + \) \( i\sin(\frac{1}{3}\tan^{-1}\sqrt{3})] \), which equals \( \sqrt[3]{2}(\cos\frac{\pi}{9} + i\sin\frac{\pi}{9}) \). And \( \cos\frac{\pi}{9} \) is what we needed to find, so this takes us in circles; hence, the futility of being able to find the third-angle identity of sine and cosine for any given angle except the known ones.

Alternate Trigonometric Solution

There is another way to represent the roots of a cubic using trigonometry. The solution is found on this Wiki Books page. It is a complicated solution that works on the full cubic equation has three real roots.

That solution is of the form \( a\cos(\frac{1}{3}\arccos\theta + \frac{2k\pi}{3}) \). The solution I presented above is of the form \( a\cos(\frac{1}{3}\arctan\theta + \frac{2k\pi}{3}) \). Both seem to work if the cubic equation has three real roots.

THE FULL CUBIC

If we know one root of the cubic \( x^{3} + Ax^{2} + Bx + C = 0 \), then we can find the other roots easily.

Suppose the roots are \( r_{1} \), \( m + ni \) and \( m - ni \). If n turns out to be 0, the other 2 roots will be real.

The sum of the roots is \( -A \). Therefore:

(i) \( r_{1} + m + ni + m - ni = -A \)

(ii) \( m = -\frac{r_{1}+A}{2} \)

The product of the roots is \( -C \). Therefore:

(iii) \( r_{1}(m+ni)(m-ni) = r_{1}(m^{2}+n^{2}) = -C \)

(iv) \( n^{2} = -\frac{(r_{1}+A)^{2}}{4} - \frac{C}{r_{1}} \)

(v) \( n = \pm \left(\sqrt{\frac{(r_{1}+A)^{2}}{4} + \frac{C}{r_{1}}}\right)i \)

Combining the result of m and n, we get:

(vi) \( -\frac{r_{1}+A}{2} \pm \left(\sqrt{\frac{(r_{1}+A)^{2}}{4} + \frac{C}{r_{1}}}\right)i\cdot i \)

(vii) \( -\frac{r_{1}+A}{2} \pm \sqrt{\frac{(r_{1}+A)^{2}}{4} + \frac{C}{r_{1}}} \)

The \( i \cdot i \) becomes -1 and the imaginary number becomes real. However, if the square root is negative, we can still end up with non-real roots. It may seem B is not invoved in this formula, but B is actually included in the derivation of \( r_{1} \).

The above formula can also be written as \( -\frac{r_{1}+A}{2} \pm \frac{1}{2}\sqrt{\frac{Ar_{1}^{2} + (A^{2}-B)r_{1}+3C}{r_{1}}} \) when substituting \( r_{1}^{3} + Ar_{1}^2 = -Br_{1} - C \).

Example 5

Find the roots of \( x^3 - 2x^2 - 5x + 6 = 0 \).

By trial and error, we can determine one root is 1. The other roots are:

(i) \( -\frac{1+(-2)}{2} \pm \sqrt{\frac{(1+(-2))^{2}}{4} + \frac{6}{1}} \)

(ii) \( \frac{1}{2} \pm \sqrt{\frac{1}{4} + 6} \)

(iii) \( \frac{1}{2} \pm \sqrt{\frac{25}{4}} \)

(iv) \( \frac{1}{2} \pm \frac{5}{2} \)

(iv) \( 3, -2 \)

The roots are -2, 1, and 3.

Example 6

Find the roots of \( x^{3} - x^{2} + x - 1 = 0 \).

This is a simple cubic equation. One root is 1 by trial and error since we always try 1 first.

The other roots are:

(i) \( -\frac{1+(-1)}{2} \pm \sqrt{\frac{(1+(-1))^{2}}{4} + \frac{-1}{1}} \)

(ii) \( 0 \pm \sqrt{0 - 1} \)

(iii) \( \pm i \)

The roots are \( 1, \, i, \, -i \).

Example 7

Find the roots of \( x^{3} + x^{2} + x - 3 = 0 \).

This is another simple cubic with at least one real root, which is 1. Let’s find the other roots using the simple formula:

(i) \( -\frac{1+1}{2} \pm \sqrt{\frac{(1+1)^{2}}{4} + \frac{-3}{1}} \)

(ii) \( -\frac{2}{2} \pm \sqrt{\frac{(2)^{2}}{4} - 3} \)

(iii) \( -1 \pm \sqrt{-2} \)

(iv) \( -1 \pm \sqrt{2}i \)

The roots are \( 1, \, -1 + \sqrt{2}i, \, -1 - \sqrt{2}i \). Let’s check to ensure the answer is correct.

\( (-1 + \sqrt{2}i)^{2} = -1 - 2\sqrt{2}i \)

\( (-1 + \sqrt{2}i)^{3} = 5 + 2\sqrt{2}i \)

\( 5 + \sqrt{2}i - 1 - 2\sqrt{2}i -1 + \sqrt{2}i - 3 = 0 \)

The math checks out.

Cube Roots vs Roots of a Cubic

Numbers have cube roots. Polynomials have roots. A real number has 3 cube roots: 1 real root and 2 non-real roots. A complex number has 3 non-real roots.

A cubic polynomial has at least 1 real root. It can have 2 roots if one is a double root.

The cube roots of a number are cyclic in nature. Therefore, when plotted on a complex plane, they are equally spaced apart on a circle that passes through the three points. When connected, an equilateral triangle is formed.

One root of a real number is real, therefore, it will lie on the x-axis - this is one of the vertices of the equilateral triangle. The other 2 non-real roots be parallel to the y-axis and will form the other 2 vertices of the equilateral triangle. On the other hand, all three roots of a non-real number are non-real. They will lie anywhere on the circle but still form an equilateral triangle.

When the cube roots of a cubic equation are plotted on a complex plane, they form an isosceles triangle. They may form an equilateral triangle in special cases where the roots of the polynomial turns out to be the roots of a number. For example, \( x^{3} - 8 = 0 \) will have roots that form an equilateral triangle, since the equation basically reduces to the cube roots of 8.

Figure 5 above shows the difference. Points B, E, and F are the cube roots of \( 2.649^{3} \). They form an equilateral triangle.

Points B, C, and D are the roots of the cubic equation \( x^{3} - 4x - 8 = 0 \). When the root is positive, the vertex of the real root will be on the positive x-axis and the other 2 non-real roots will be in the 2nd and 3rd quadrant.

Remember the formula \( x = -\frac{1}{2}r_{1} \pm\frac{1}{2}\sqrt{\frac{3C + Br_{1}}{r_{1}}}i \) for the other roots of a cubic equation? If the other roots are non-real, then \( -\frac{1}{2}r_{1} \) represents the real part of the complex number and it is the same for both non-real roots. Hence, we have an isosceles triangle for the roots of a cubic equation.

THE EQUILATERAL TRIANGLE

Is there a cubic equation of the form \( x^{3} + Bx = C \) which yields all real rational roots?

We looked at a couple of examples where one root is rational but the other two are irrational. I have not come across an equation yet. Of course, if the equation is of the form \( x^{3} + Bx^{2} +Cx + D = 0 \), then this is possible.

I played around with some numbers and found that \( x^{3} - 7x = 6 \) has roots -2, -1, and 3. Using the formula gives the first root as \( \sqrt[3]{3 + \frac{10\sqrt{3}}{9}i} + \sqrt[3]{3 - \frac{10\sqrt{3}}{9}i} \), which equals 3.

Converted to trigonometric form, the solution is \( 2\sqrt{\frac{7}{3}}\cos(\frac{1}{3}\tan^{-1}\frac{10\sqrt{3}}{27}) \).

But then I realized, the answer is much simpler. If the sum of three numbers equal 0, then their cubic equation will be \( x^{3} + Bx = C \), since the \( x^{2} \) coefficient is 0.

Interestingly, the real roots of a cubic equation \( x^{3} + Bx = C \) has a connection with the equilateral triangle.

Link to the Equilateral Triangle

If three numbers add to 0 and they are plotted on a polar graph 120 degrees apart, they form an equilateral triangle. These 3 numbers also represent the roots of the cubic equation \( x^{3} + Bx = C \).

In Figure 5a, the numbers 5, 3, and -8 have been plotted 120 degrees apart. The coordinate (8; 357.3°) has a negative radius, hence it shows as positive in Geogebra. The equation of this cubic is \( x^{3} - 49x + 120 = 0 \). The length of the side of the equilateral triangle is 7, which is \( \sqrt{|B|} \), which we will learn later.

You can see the three numbers form an equilateral triangle. The circle that passes through the 3 points also passes through the origin. Hence we have a cyclic quadrilateral. One side of the equilateral triangle is a diagonal of the cyclic quadrilateral. If we let \( s \) be the length of the side of the equilateral triangle and \( r_{1} \), \( r_{2} \), and \( r_{3} \) be the roots (represented by the distance from the pole to the 3 vertices of the equilateral triangle), then by Ptolemy’s theorem, we have \( r_{1}s + r_{2}s = r_{3}s \), proving the quadrilateral is cyclic. (This property is well known.) Hence, \( r_{1} + r_{2} = r_{3} \) or \( r_{1} + r_{2} - r_{3} = 0 \).

The Polar Cooridnates of Roots

When a cubic has a complex root, it can be plotted on a complex plane as a coordinate pair. However, when the roots are all real, it is just a number, not a coordinate pair. However, when the real roots are represented in its trigonometric form, we have a polar pair.

Consider the roots of \( x^{3} - 2x + 1 = 0 \) we got above. The first root was \( -\frac{2\sqrt{6}}{3}\cos\left( \frac{1}{3}\tan^{-1}{\frac{\sqrt{15}}{9}}\right) \). Let the radius (the distance from the pole) be \( r= -\frac{2\sqrt{6}}{3}\cos\left( \frac{1}{3}\tan^{-1}{\frac{\sqrt{15}}{9}}\right) \). Let the angle be \( \frac{1}{3}\tan^{-1}{\frac{\sqrt{15}}{9}} \). This angle is actually irrelevant as long as we choose 3 angles that are 120 degrees apart.

Now, let’s do the same for the other 2 roots. Then, let’s plot these 3 roots on a polar plot. Finally, connect the three roots with segments and find the angle between them. Lo and behold! The equilateral triangle emerges again, just like above.

The length of the sides of the equilateral triangle is equal to \( \sqrt{3}(\sqrt{a^{2}+b^{2}})^{1/3} \). Essentially, \( a^{2} = \frac{C^{2}}{4} \) and \( b^{2} = \frac{|12B^{3} + 81C^{2}|}{18^{2}} \), so the length is also equal to \( \sqrt{3}\left(\sqrt{\frac{C^{2}}{4} + \frac{|12B^{3} + 81C^{2}|}{18^{2}}}\right)^{1/3} \) in terms of the coefficients of the cubic equation.

Note: In Figure 6, negative roots are shown as positive when plotted on polar graph, with the angle that is 180 degrees greater than the original angle.

Figure 6 shows the cubic equation \( x^{2} - 2x + 1 \). A circle passing through the polar points passes through the 3 roots of the cubic equation. The length of the side of the equilateral triangle is \( \sqrt{2} \). This is derived from \( \sqrt{3}\left(\sqrt{\frac{C^{2}}{4} + \frac{|12B^3 + 81C^2|}{18^{2}}}\right)^{1/3} \, = \) \( \sqrt{3}\left(\sqrt{\frac{1^{2}}{4} + \frac{|12(-2)^3 + 81(1)^2|}{18^{2}}}\right)^{1/3} \, = \) \( \sqrt{3}\left(\sqrt{\frac{1}{4} + \frac{|-5|}{108}}\right)^{1/3} \, = \) \( \sqrt{3}\left( \sqrt{\frac{128}{432}}\right)^{1/3} \, =\) \( \sqrt{3}\left(\sqrt{\frac{8}{27}}\right)^{1/3} \, = \) \( \sqrt{3}\sqrt{\frac{2}{3}} \, = \) \( \sqrt{2} \).

When B is negative, \( \sqrt{3}\left(\sqrt{\frac{C^{2}}{4} + \frac{|12B^3 + 81C^2|}{18^{2}}}\right)^{1/3} \, = \sqrt{|B|} \,\) because \( |12B^{3} + 81C^{2}| \) becomes \( 12B^{3} - 81C^{2} \). The reader can confirm this by putting the \( \sqrt{3} \) inside the radical and evaluating the result. So the length of the equilateral triangle is \( \sqrt{|B|} \, \)! If \( B \gt 0 \), then the equation will only have 1 root, so we cannot form an equilateral triangle.

Figures 7 and 8 show 2 more examples we have seen before. You can confirm the length of the equilateral triangle is \( \sqrt{|B|} \).

UNSOLVED PROBLEMS

Problem 1

When we use the cubic formula and obtain a real root expressed as cube roots of complex numbers like \( \sqrt[3]{a+bi} + \sqrt[3]{a-bi} \), then the trigonometric form of the solution is \( 2(\sqrt{a^{2}+b^{2}})^{1/3}\left[ \cos (\frac{1}{3}\tan^{-1}(\frac{b}{a})) \right] \). However, the other solutions seem to be \( 2(\sqrt{a^{2}+b^{2}})^{1/3}\left[ \cos (\frac{1}{3}\tan^{-1}(\frac{b}{a}) + \frac{2\pi}{3}) \right] \) and \( 2(\sqrt{a^{2}+b^{2}})^{1/3}\left[ \cos (\frac{1}{3}\tan^{-1}(\frac{b}{a})+ \frac{4\pi}{3}) \right] \).

I don’t have a formal proof of this, except the observation that these three roots add to 0, which is correct for cubic equation of the form \( x^{3} + Bx = C \).

Consider the cubic equation \( x^{3} - 2x + 1 = 0 \). The solution is \( x = -\sqrt[3]{\frac{1}{2} + \frac{\sqrt{15}}{18}i} \, - \) \( \sqrt[3]{\frac{1}{2} - \frac{\sqrt{15}}{18}i} \).

In trigonometric form, this is \( -\frac{2\sqrt{6}}{3}\cos\left( \frac{1}{3}\tan^{-1}{\frac{\sqrt{15}}{9}}\right) \). This value is the negative value of the golden ratio: \( -\frac{1+\sqrt{5}}{2} \approx -1.618 \).

The second root is: \( -\frac{2\sqrt{6}}{3}\cos\left( \frac{1}{3}\tan^{-1}{\frac{\sqrt{15}}{9}} + \frac{2\pi}{3}\right) \, = 1\).

The third root is: \( -\frac{2\sqrt{6}}{3}\cos\left( \frac{1}{3}\tan^{-1}{\frac{\sqrt{15}}{9}} + \frac{4\pi}{3}\right) \, = \frac{-1+\sqrt{5}}{2} \approx 0.618 \). This is the inverse of the golden ratio.

Proof

We can at least prove the sum of the three trigonometric roots adds to 0. We simply need to show that \( a\cos(\theta) + a\cos(\theta + \frac{2\pi}{3}) \, + \) \( a\cos(\theta + \frac{4\pi}{3}) = 0 \), for some a and angle θ because the roots are of this form. Let’s expand using the sum of angles identity of cosine.

(i) \( a\cos(\theta) + a\cos(\theta)\cos(\frac{2\pi}{3}) - a\sin(\theta)\sin(\frac{2\pi}{3}) \, + \) \( a\cos(\theta)\cos(\frac{4\pi}{3}) - a\sin(\theta)\sin(\frac{4\pi}{3}) \)

(ii) \( a\cos(\theta) - \frac{1}{2}a\cos(\theta) - \frac{\sqrt{3}}{2}a\sin(\theta) \, - \) \( \frac{1}{2}a\cos(\theta) + \frac{\sqrt{3}}{2}a\sin(\theta) \)

(iii) \( 0 \)

Therefore, the sum of \( 2(\sqrt{a^{2}+b^{2}})^{1/3}\left[ \cos (\frac{1}{3}\tan^{-1}(\frac{b}{a})) \right] \), \( 2(\sqrt{a^{2}+b^{2}})^{1/3}\left[ \cos (\frac{1}{3}\tan^{-1}(\frac{b}{a}) + \frac{2\pi}{3}) \right] \), and \( 2(\sqrt{a^{2}+b^{2}})^{1/3}\left[ \cos (\frac{1}{3}\tan^{-1}(\frac{b}{a})+ \frac{4\pi}{3}) \right] \) has to be 0 since they have the same coefficient and the same angle.

Problem 2

Is there a way to figure out why \( 2\sqrt{5}\cos(\frac{1}{3}\tan^{-1}\frac{11}{2}) \) equals 4? Working backward, we can see it does, but is there a method to figure this out working forward? If so, we can apply it to other expressions as well.

Anirdesh.com Last Updated: August 26 2025, 14:43:34.