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a+b+c=0 then find the value of a2/(a2-bc)+b2/(b2-ca)+c2/(c2-ab)

a+b+c=0 then find the value of a2/(a2-bc)+b2/(b2-ca)+c2/(c2-ab)

Grade:12th pass

2 Answers

Ashams Thomas
13 Points
5 years ago
From a+b+c=0, a+b=-c
Substituting for -c,
a2/[a2+ b*(a+b)]+ b2/[b2 +a*(a+b)]+ (a+b)2/[(a+b)2-ab]
= a2/(a2+ab+b2) + b2/(a2+ab+b2) + a2+2ab+b2/(a2+ab+b2)
Nawal Kishore Prasad
15 Points
4 years ago
 
From a+b+c=0, a+b=-c
Substituting for -c,
a2/[a2+ b*(a+b)]+ b2/[b2 +a*(a+b)]+ (a+b)2/[(a+b)2-ab]
= a2/(a2+ab+b2) + b2/(a2+ab+b2) + a2+2ab+b2/(a2+ab+b2)
=2(a^2+b^2+ab)/(a^2+b^2+c^2)
=2

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