Annoying Fractions

If $\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=0$ Then $\frac{a^2}{b+c}+a+\frac{b^2}{a+c}+b+\frac{c^2}{a+b}+c=a+b+c$ $\frac{a(a+b+c)}{b+c}+\frac{b(a+b+c)}{a+c}+\frac{c(a+b+c)}{a+b}=a+b+c$ $(a+b+c)(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b})=a+b+c$ So $a+b+c=0\text{ or }\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=1$ If $a+b+c=0$ ,then $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=-3$ Hence, $(-3)+1=-2$

By given condition, $\sum_{ cyc } a^2(a+c)(a+b) = 0$

Therefore $(a+b+c)( a^3 + b^3 + c^3 + abc ) = 0$

i) $a+b+c=0$

$\sum_{ cyc } \frac{a}{b+c} = - 3$

ii) $a^3 + b^3 + c^3 + abc = 0$

$\sum_{ cyc } \frac{a}{b+c} = \frac{ \sum_{ cyc } a(a+c)(a+b) } { (a+b)(b+c)(c+a) }$

= $\frac{ \sum_{ cyc } ( a^3 + a^2 (b+c) +abc ) } { (a+b)(b+c)(c+a) }$

= $\frac{ -abc + a^2 (b+c) + b^2 (c+a) + c^2 (a+b) + 3abc } { (a+b)(b+c)(c+a) }$

= $\frac{ (a+b)(b+c)(c+a) } { (a+b)(b+c)(c+a) } = 1$

By i) & ii) , Sol) = $-2$