Problem
ALG-B1-M01-P008 Zero Sum
#8
★★☆☆☆ Level 2 of 5
Let \(a+b+c=0\). Prove that \(a^2+b^2+c^2=-2(ab+bc+ca)\) and \(a^3+b^3+c^3=3abc\).
Square \(a+b+c=0\), then use the cubic identity.
From \((a+b+c)^2=0\), we get \(a^2+b^2+c^2+2ab+2bc+2ca=0\). Hence \(a^2+b^2+c^2=-2(ab+bc+ca)\).
Also \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0\). Thus \(a^3+b^3+c^3=3abc\).
The two formulas are often used together.