Problem
ALG-B1-M01-P013 Equality in the Cubic Identity
#13
★★★☆☆ Level 3 of 5
Prove that for real \(a,b,c\), if \(a+b+c>0\) and \(a^3+b^3+c^3=3abc\), then \(a=b=c\).
Move \(3abc\) to the left and factor.
By the identity, \[ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca). \] The left side is zero, and \(a+b+c>0\), so the second factor is zero.
But \[ 2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2. \] A sum of squares is zero only when \(a=b=c\).
Here the identity gives not just factorization, but an equality condition.