Problem
ALG-B1-M06-P007 Differences of three numbers
#7
★★☆☆☆ Level 2 of 5
Let \(a+b+c=6\) and \(ab+bc+ca=11\). Find \((a-b)^2+(b-c)^2+(c-a)^2\).
First find \(a^2+b^2+c^2\).
From \((a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\) we get \(36=a^2+b^2+c^2+22\), so \(a^2+b^2+c^2=14\).
Then \((a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2-ab-bc-ca)=2(14-11)=6\).
A useful formula for later symmetry problems.