Problem
ALG-B1-M02-P013 When the Cubic Expression Is Zero
#13
★★★☆☆ Level 3 of 5
Let \(a+b+c>0\) and \(a^3+b^3+c^3=3abc\). Prove that \(a=b=c\).
Factor \(a^3+b^3+c^3-3abc\) and use a sum of squares.
The condition gives \(a^3+b^3+c^3-3abc=0\). By the identity, this equals \((a+b+c)(a^2+b^2+c^2-ab-bc-ca)\).
Since \(a+b+c>0\), the second factor is zero. But \(2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2\). Hence \(a=b=c\).
The problem connects factorisation with an equality case.