Question

If a, b, c are rational numbers such that $a^2+b^2+c^2-ab-bc-ca=0$, prove that $a=b=c$.

Answer 1

$a,b,cϵQ$ s.t.

$a^2+b^2+c^2-ab-bc-ca=0$

To prove $a=b=c$

$2(a^2+b^2+c^2-ab-bc-ca)=0$

$2a^2+2b^2+2c^2-2ab-2bc-2ca=0$

$a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0$

$(a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0$

Is is possible only when

$a-b=0\Rightarrow a=b$

$b-c=0\Rightarrow b=c$

$c-a=0\Rightarrow c=a$

$\Rightarrow a=b=c$ proved.