Question

If a + b + c = 0, then the value of $\frac{a^2+b^2+c^2}{a^2-bc}$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is C 2

Given, $a+b+c=0$

$a=-b-c-\left(1\right)$

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

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

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

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

$⟹2(a+b+c)=K(ab+ca+bc)$

$⟹K=2$