<s>[INST] Cubic roots of the equation $x^3-x-2=0$ If $\alpha,\beta,\gamma$ are the cubic roots of the equation $x^3-x-2=0$, then find the value of $\alpha^5+\beta^5+\gamma^5$.
One of the root is real and two roots are imaginary.
I get three equation.
$\alpha+\beta+\gamma=0$;
$\alpha\beta+\beta\gamma+\gamma\alpha=-1$
$\alpha\beta\gamma=2$
But not able to get the requisite result.
 [/INST] Note:
$$a^3=a+2 \Rightarrow a^5=a^3+2a^2=2a^2+a+2$$
Hence:
$$a^5+b^5+c^5=2(a^2+b^2+c^2)+(a+b+c)+6=\\
2((a+b+c)^2-2(ab+bc+ca))+6=\\
2(0^2-2(-1))+6=10.$$
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