Question

If $ab+bc+ca=0$ , then the value of $\frac{1}{a^2-bc}+\frac{1}{b^2-ca}+\frac{1}{c^2-ab}$ will be

• A. -1
• B. a+b+c
• C. 0
• D. ab

Answer 1

The correct option is B a+b+c

$\frac{1}{a^2-bc}+\frac{1}{b^2-ca}+\frac{1}{c^2-ab}$

Given $ab+bc+ca=0$

now $-bc=ab+ca$

$-ca=ab+bc$

$-ab=bc+ca$

$\frac{1}{a^2+(ab+ca)}+\frac{1}{b^2+(ab+bc)}+\frac{1}{c^2+(bc+ca)}$

$=\frac{1}{a(a+b+c)}+\frac{1}{b(a+b+c)}+\frac{1}{c(a+b+c)}$

$=\frac{bc+ca+ab}{abc(a+b+c)}=0$.