Sum of the cubes of the roots in a cubic

Question

I had an argument with my brother this morning about finding p^3+q^3+s^3 where p and q and s are the roots of a cubic. What is this value expressed in terms of the coefficients of the cubic?

(Sorry if this was rushed, but I want to win the argument)

Answer 1

Problem: Let the cubic polynomial be $ax^3 + bx^2 + cx + d$ and $p, q, r$ be it’s roots.

To find: $p^3 + q^3 + r^3$

We know the following identity: $p^3 + q^3 + r^3 - 3pqr = (p+q+r)(p^2 + q^2 + r^2 - pq - qr - pr)$

By Vieta’s relations: $$p + q + r = \frac{-b}{a}\\ pq + qr + pr = \frac{c}{a}\\ pqr = \frac{-d}{a}\\$$

The only thing left to find is $p^2 + q^2 + r^2: $

By using the following identity $(p + q + r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + pr)$ we get: $$\frac{b^2}{a^2} = p^2 + q^2 + r^2 + \frac{2c}{a} \\ \frac{b^2}{a^2} - \frac{2c}{a} = p^2 + q^2 + r^2 \implies \frac{b^2 - 2ac}{a^2}$$

By substituting these in the original identity as stated before: $$p^3 + q^3 + r^3 - \frac{3d}{a} = \Bigg(\frac{-b}{a}\Bigg) \Bigg(\frac{b^2 - 2ac}{a^2} - \frac{c}{a}\Bigg)\\ p^3 + q^3 + r^3 = \Bigg(\frac{-b}{a}\Bigg) \Bigg(\frac{b^2 - 3ac}{a^2}\Bigg) - \frac{3d}{a} \implies \frac{-b^3 - 3a^2d + 3abc }{a^3}$$

Edit: Thank you Ayan. $\tag*{$\blacksquare$}$