How do I find the sum of the cubes of the roots in a cubic polynomial?

Question

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?

Answer 1

Try expanding $(p+q+r)^3$ and $(pq+pr+qr)(p+q+r)$ and see if anything comes to mind. You should be able to retrieve $p^3+q^3+r^3$ from there.

Answer 2

Hint: Note that $$p^3+q^3+r^3-(p^2+q^2+r^2)+(p+q+r)=6$$ and if; $$p+q+r= 1\implies (p+q+r)^2=1\implies p^{2}+q^{2}+r^{2}+2(pq+pr+qr)=1$$

Answer 3

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

$$p^3 + q^3 + r^3 = 1^2 - 2(1) - 1 + 6 = 4$$