Problem
ALG-B1-M06-P002 Two identities with zero sum
#2
★☆☆☆☆ Level 1 of 5
Let \(a+b+c=0\). Prove that \(a^2+b^2+c^2=-2(ab+bc+ca)\) and \(a^3+b^3+c^3=3abc\).
The first identity comes from squaring the sum. For the second, use the factorization of \(a^3+b^3+c^3-3abc\).
From \((a+b+c)^2=0\) we get \(a^2+b^2+c^2+2(ab+bc+ca)=0\), which gives the first equality.
Also, \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\). The right-hand side is zero, hence \(a^3+b^3+c^3=3abc\).
These two identities will be used repeatedly in harder problems.