Question

If $a^2+b^2+c^2-ab-bc-ca=0$ then prove that $a=b=c.$

Answer 1

Solve for the required proof

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

$\Rightarrow a^2+b^2+c^2=ab+bc+ca$
On multiplying both sides of the equation with $2$, we get
$(a^2+b^2+c^2)=2(ab+bc+ca)$

$\Rightarrow$ $2a^2+2b^2+2c^2=2ab+2bc+2ca$

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

$\Rightarrow \left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0$

$\Rightarrow$ ${\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2=0\left[\because x^2-2xy+y^2={\left(x-y\right)}^2\right]$

$\Rightarrow$ $\left(a-b\right)=\left(b-c\right)=\left(c-a\right)=0\left[\because x^2+y^2+z^2=0\Rightarrow x=y=z=0\right]$

Consider $\left(a-b\right)=0\Rightarrow a=b$

Similarly by considering $\left(b-c\right)=0,\left(c-a\right)=0$ we get $b=c,c=a$ respectively

Thus we can say that $a=b=c$

Hence proved