Question

If `a+b+c=0`, `ab+bc+ca=2` and `abc=-1` then what is the value of `a^2/(b+c) + b^2/(c+a) + c^2/(a+b)` ?

Answer 1

`a^2/(b+c)+b^2/(c+a)+c^2/(a+b)=0`

Explanation:

As `a+b+c=0`, we have `a+b=-c`, `b+c=-a` and `c+a=-b`.

Hence `a^2/(b+c)+b^2/(c+a)+c^2/(a+b)`

= `(a^2(c+a)(a+b)+b^2(b+c)(a+b)+c^2(b+c)(c+a))/((a+b)(b+c)(c+a))`

= `(a^2bc+b^2ac+c^2ab)/(-abc)`

= `-(a+b+c)`

= `0`