Question

ab+bc+ca=0
1/a²-bc + 1/b²-ca + 1/c²-ab​

Answer 1

$\sf \: ab+bc+ca=0$

$\sf \: bc=−ca−ab$

$\sf \: bc=−a(b+c)---eq1$

like wise,

$\sf \: ab=−c(a+b)---eq2$

$\sf \: and \: ca=−b(c+a)---eq3$

$\boxed{ \pink{ \sf \: \frac{1}{a {}^{2} - bc} + \frac{1}{b {}^{2} - bc} + \frac{1}{c {}^{2} - ab } }}$

Taking the value from eq1, 2&3

$\bf{ \sf \: \frac{1}{a {}^{2} + a(a + b + c) } + \frac{1}{b {}^{2} + b (a + b + c)} + \frac{1}{c {}^{2} + c(a + b + c)} }$

$\boxed{ \sf \: \frac{1}{a(a + b + c)} + \frac{1}{b (a + b + c)} + \frac{1}{c(a + b + c)}}$

Ab+Bc+Ca=0

Hence, This is Answer.