The polynomial $p(x) = x^3 - 2x + 2$ has roots $\alpha$, $\beta$ and $\gamma$ - HSC - SSCE Mathematics Extension 2 - Question 5 - 2017 - Paper 1

Again using Vieta's formulas, the sum of the products of the roots taken two at a time is:

$\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{-2}{1} = -2$

The product of the roots is given by:

$\alpha\beta\gamma = -\frac{d}{a} = -\frac{2}{1} = -2$

We can use the identity for the sum of cubes:

$\alpha^3 + \beta^3 + \gamma^3 = 3\alpha\beta\gamma + (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2)$

Now, we can express $\alpha^2 + \beta^2 + \gamma^2$ as:

$\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)$

Substituting the known values:

$\alpha^2 + \beta^2 + \gamma^2 = (0)^2 - 2(-2) = 4$

Thus:

$\alpha^3 + \beta^3 + \gamma^3 = 3(-2) + 0(4) = -6$