Cubic roots of the equation $x^3-x-2=0$

Question

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.

Answer 1

Note: $$a^3=a+2 \Rightarrow a^5=a^3+2a^2=2a^2+a+2$$ Hence: $$a^5+b^5+c^5=2(a^2+b^2+c^2)+(a+b+c)+6=\\ 2((a+b+c)^2-2(ab+bc+ca))+6=\\ 2(0^2-2(-1))+6=10.$$

Answer 2

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:

$$0^2 = (a + b + c)^2 = (a^2 + b^2 + c^2) + 2(ab + ac + bc)$$

$$= (a^2 + b^2 + c^2) + 2(-1) \implies $$

$$(a^2 + b^2 + c^2) = 2.$$

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So, now you can compute

$$a^5 + b^5 + c^5$$

$$= (a^3 \times a^2) + (b^3 \times b^2) + (c^3 \times c^2)$$

$$= [(a + 2) \times a^2] + [(b + 2) \times b^2] + [(c + 2) \times c^2]$$

$$= (a^3 + b^3 + c^3) + 2(a^2 + b^2 + c^2)$$

$$= [(a + 2) + (b + 2) + (c + 2)] + 2(2)$$

$$= [0 + 6] + 4 = 10.$$