Question

If $a+b+c=0$, show that $a^2-bc=\left(\frac{a^2+b^2+c^2}{2}\right)$

Answer 1

Given:

$a+b+c=0$ ....$\left(1\right)$

Squaring both the sides, we get

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

$\Rightarrow a^2+b^2+c^2+2(ab+bc+ac)=0$

$\Rightarrow a^2+b^2+c^2=-2[bc+a(b+c\left)\right]$

$\Rightarrow a^2+b^2+c^2=-2[bc+a(-a\left)\right]$ ....$[$From $\left(1\right)]$

$\Rightarrow a^2+b^2+c^2=-2(bc-a^2)$

$\Rightarrow a^2-bc=\frac{a^2+b^2+c^2}{2}$

Hence proved