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$a,b$ and $c$ are the roots of

$$x^3-x-1=0$$

Find

$$a^{\frac{2}{3}}+b^{\frac{2}{3}}+c^{\frac{2}{3}}$$

To solve this question I called the quantity A and I calculated $A^3$. I have a feeling there is a better way.

Question From Jalil Hajimir

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Kian
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    Since $x^3 - x - 1= 0$ has two imaginary roots, the cubic roots are not well-defined, since there are three of them, all imaginary. – WhatsUp Dec 04 '19 at 22:18

1 Answers1

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There are one real root $a$ and a pair of imaginary ones, $b$ and $c$. Per $abc = 1$, express $b=\frac1{a^{1/2}}e^{i \theta} $ and $c=\frac1{a^{1/2}}e^{-i \theta} $. Substitute $b$ and $c$ into $x^3-x-1=0$

$$\frac1{a^{3/2}}e^{\pm i 3\theta} -\frac1{a^{1/2}}e^{\pm i \theta} -1=0$$

The difference of the two equations leads to

$$\frac1{a^{3/2}}\sin3\theta-\frac1{a^{1/2}}\sin\theta=0 \implies \cos2\theta = \frac{a-1}2$$

Then

\begin{align} a^{\frac{2}{3}}+b^{\frac{2}{3}}+c^{\frac{2}{3}} =&\ a^{\frac{2}{3}} + a^{-\frac{1}{3}}(e^{ i \frac{2\theta}{3}}+e^{- i \frac{2\theta}{3}})\\ =&\ a^{\frac{2}{3}} + 2a^{-\frac{1}{3}}\cos\left(\frac{1}{3}\cos^{-1}\frac{a-1}2\right) \end{align}

where the real root $a$ is given analytically by

$$a = \sqrt[3]{\frac12 -\frac{\sqrt{69}}{18}} + \sqrt[3]{\frac12 +\frac{\sqrt{69}}{18}} $$

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