Question

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

• A. $1$
• B. $0$
• C. $\frac{1}{abc}$
• D. $abc$

Answer 1

The correct option is B $0$

$ab+bc+ca=0$

$ab+ca=-bc$ (1)

$ab+bc=-ca$ (2)

$bc+ca=-ab$ (3)

$\frac{1}{a^2-bc}+\frac{1}{b^2-ca}+\frac{1}{c^2-ab}$ (4)

Substituting (1),(2) and (3) in (4)

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

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

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

Hence the correct option is (B).