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.

Alternative approach that is very similar to the answer of farruhota.
Suppose that the roots are $a,b,c$.
You know that each value of $a,b,c$ satisfies
$$x^3 = x + 2.$$
From the coefficient of the $x^2$ term, you also know that
$$(a + b + c) = 0.$$
Also, from the coefficient of the $x^1$ term, you know that
$$ab + ac + bc = -1.$$
This allows you to compute $(a^2 + b^2 + c^2)$, as follows:
$$0^2 = (a + b + c)^2 = (a^2 + b^2 + c^2) + 2(ab + ac + bc)$$
$$= (a^2 + b^2 + c^2) + 2(-1) \implies $$
$$(a^2 + b^2 + c^2) = 2.$$

So, now you can compute
$$a^5 + b^5 + c^5$$
$$= (a^3 \times a^2) + (b^3 \times b^2) + (c^3 \times c^2)$$
$$= [(a + 2) \times a^2] + [(b + 2) \times b^2] + [(c + 2) \times c^2]$$
$$= (a^3 + b^3 + c^3) + 2(a^2 + b^2 + c^2)$$
$$= [(a + 2) + (b + 2) + (c + 2)] + 2(2)$$
$$= [0 + 6] + 4 = 10.$$
