Hello !

See,

a + b + c = 0

a= - b - c ............................ 1

b = - a - c ............................ 2

c = - b - a ............................ 3

On solving the first term i . e 1/(b²+c²-a²)

in this on substituting the value of a using eqn 1 we get

⇒ 1/(b²+c²-(-b-c)²)

⇒1/(b²+c²-(b²+c²+2bc))

⇒1/(b²+c²-b²-c²-2bc) = **1/(-2bc) ** the first term.

Now on solving the second term i.e 1/(c²+a²-b²)

on substituting the value of b using eqn 2 we get

⇒1/(c²+a²-a²-c²-2ac) = **1/(-2ac)** the second term

Now again on solving the third term i.e. 1/(a²+b²-c²)

on substituting the value of c using eqn 3 we get

⇒1/(a²+b²-a²-b²-2ab) = ** 1/(-2ab)** the third term

Now adding all the three terms together we get

**⇒ 1/(-2bc) + 1/(-2ac) + 1/(-2ab) = -1/(2bc) - 1/(2ac****) - 1/(2ab)**

**⇒ (-a-b-c)/2abc = -(a+b+c)/2abc = - 0/2abc** (as a+b+c =0)

⇒ __0 Ans.__

The answer is option (e) in the circled question !.

I hope the answer is correct !

Please mark it as the best if you are satisfied!

Thank you !

Have a nice day !