<s>[INST] $a,b$ and $c$ are roots of $x^3-x-1=0$. Find $a^{\frac{2}{3}}+b^{\frac{2}{3}}+c^{\frac{2}{3}}$ $a,b$ 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

 [/INST] There are one real and a pair of imaginary roots. Let $a$ be the real root available analytically and express $b=\frac1{a^{1/2}}e^{i \theta} $ and $c=\frac1{a^{1/2}}e^{-i \theta} $ based on $abc = 1$. 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,
$$a^{\frac{2}{3}}+b^{\frac{2}{3}}+c^{\frac{2}{3}}
=a^{\frac{2}{3}} + a^{-\frac{1}{3}}(e^{ i \frac{2\theta}{3}}+e^{- i \frac{2\theta}{3}})=a^{\frac{2}{3}} + 2a^{-\frac{1}{3}}\cos\left(\frac{1}{3}\cos^{-1}\frac{a-1}2\right)$$
where the real root $a$ is given by,
$$a = \sqrt[3]{\frac12 -\frac{\sqrt{69}}{18}} + \sqrt[3]{\frac12 +\frac{\sqrt{69}}{18}} $$
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