<s>[INST] How do I find the sum of the cubes of the roots in a cubic polynomial? I have an equation, $x^3-x^2+x-2$, with three distinct roots, $p$, $q$ and $r$. What is the value of $p^3+q^3+r^3$? 
I'm not sure how to do this. Using Vieta's formula, we know that:
$pqr= 2$
$pq+pr+qr= 1$
$p+q+r= 1$
After this, what should I do? 
 [/INST] Since p, q and r are solutions to the polynomial
$$x^3 - x^2 + x - 2 = 0,$$  
i.e. $$ x^3 = x^2 - x + 2,$$
we have the following equations:
$$p^3 = p^2 - p + 2$$
$$q^3 = q^2 - q + 2$$
$$r^3 = r^2 - r + 2 .$$
Summing these equations yields
$$p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6.$$
Next we make use of the following identity
$$p^2 + q^2 + r^2  = (p + q + r)^2 - 2(pq+pr+qr).$$
Substiting the above produces
$$p^3 + q^3 + r^3 = (p + q + r)^2 - 2(pq + pr + qr) - (p + q + r) + 6,$$
that is
$$p^3 + q^3 + r^3 = 1^2 - 2(1) - 1 + 6 = 4$$
</s>