Symmetric roots of polynomial

Question

Let $\alpha_1, \alpha_2, \alpha_3$ be the roots of the polynomial $x^3 - x^2 + 2x - 3$ $\in \mathbb{C}[x]$. Calculate $\alpha_1^3 + \alpha_2^3 + \alpha_3^3$.

What to do here exactly? I already calculated the elementary symmetric polynomial identity for $\alpha_1^3 + \alpha_2^3 + \alpha_3^3$ = $(\alpha_1 + \alpha_2 + \alpha_3)^3 - 3(\alpha_1 + \alpha_2 + \alpha_3)(\alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_2\alpha_3) - 6(\alpha_1\alpha_2\alpha_3)$.

Answer 1

let $f(x) = x^3+ b_2x^2+b_1x+b_0$

If $a_1,a_2,a_3$ - roots of f(x) then
$a_1a_2a_3=-b_0$ ,

$a_1a_2+a_1a_3+a_2a_3=b_1$ and

$a_1+a_2+a_3=-b_2$

We can get this from $f(x) = x^3+ b_2x^2+b_1x+b_0 = (x-a_1)(x-a_2)(x-a_3) = x^3 - x^2 (a_1+a_2+a_3)+ x(a_1a_2+a_1a_3+a_2a_3)-a_1a_2a_3 $

Answer 2

Let $x^3=y$

$$(x^3-3)^3=(x^2-2x)^3$$

$$(y-3)^3=(x^3)^2-2^3(x^3)-3\cdot2\cdot(x^3)(x^3-3)$$

$$(y-3)^3=(y)^2-2^3(y)-3\cdot2\cdot(y)(y-3)$$

can you please rearrange & use Vieta's formula