# Roots Of Monic Cubic

I'm currently preparing for the USA Mathematical Talent Search competition. I've been brushing up my proof-writing skills for several weeks now, but one area that I have not been formally taught about (or really self-studied) for that matter, is general polynomials beyond quadratics. In particular, I've been having trouble with the following question:

Let $r_1, r_2$, and $r_3$ be distinct roots of the monic cubic equation $P(x) :=x^3+bx^2+cx+d=0$. Prove that $r_1r_2 + r_1r_3 + r_2r_3 = c$.

I started by attempting to equate the roots $P(r_1) = P(r_2) = P(r_3) = 0$ and simplify, however this seemed like a wrong approach from the start and looked far messier than I'd expect from an introductory question on direct proofs. How would one go about solving this and, in general, problems on identities involving the roots of a polynomial?

Hint: $$x^3 + bx^2 + cx + d = (x-r_1)(x-r_2)(x-r_3)$$
• @theage This comes up all the times in contest problems. See Vieta's formulas. Here is a related problem: Let $p,q,r$ be the three roots of $x^3 + 4x^2 -4x + 1$. Find $p^3+q^3+r^3$. – MJD Jul 21 '14 at 16:20
Just use $P(x)=(x-r_1)(x-r_2)(x-r_3)$. Then compare the coefficients on each side.