value of algebric expression

Question

If $a,b,c$ are three distinct complex number and $\displaystyle \frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0$ . Then $\displaystyle \frac{a^2}{(b-c)^2}+\frac{b^2}{(c-a)^2}+\frac{c^2}{(a-b)^2}=$

what i have try

$\displaystyle \bigg(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\bigg)^2=\sum_{cyc}\frac{a^2}{(b-c)^2}+2\frac{ab(a-b)+bc(b-c)+ca(c-a)}{(a-b)(b-c)(c-a)}$

$\displaystyle \sum_{cyc} \frac{a^2}{(b-c)^2}=-2\frac{ab(a-b)+bc(b-c)+ca(c-a)}{(a-b)(b-c)(c-a)}$

How do I find value of right side expression?

Answer 1

You can use the following factoring. $$\sum_{cyc}ab(a-b)=\sum_{cyc}(a^2b-a^2c)=(a-b)(a-c)(b-c).$$

Answer 2

You can check that the numerator and the denominator are equal but of different sign by simply distributing all the products. You can also work taking common factors in the numerator in several steps, until you reach the expression in the denominator. So, taking account of the minus sign affecting the whole expression, the answer would then be $1$.

Also, while the notation $$\sum \frac{a^2}{(b-c)^2}$$ is an understandable shorthand in this context, it is of course mathematically wrong and even misleading if it got out of context.