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 C $0$

$ab+bc+ca=0----\left(1\right)$

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

$=\frac{1}{a^2+ab+ca}+\frac{1}{b^2+ab+bc}+\frac{1}{c^2+aa+ca}$ (From $\left(1\right)$

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

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

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

$=0$ From $\left(1\right)$