Answers

2016-02-17T14:53:14+05:30
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 !
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0
also go through the video on YouTube ....