Being given a cyclic summation on 3 letters equal to $1$, deduce the value of another cyclic summation

Question

If $$ \dfrac{a}{b+c} + \dfrac{b}{a+c} + \dfrac{c}{a+b} = 1 $$ then $$ \dfrac{a^2}{b+c} + \dfrac{b^2}{a+c} + \dfrac{c^2}{a+b} = \;? $$

I tried to manipulated the equation above using some properties as $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+ac+bc)$ or $$ \dfrac{a(a+c)(a+b)+b(b+c)(a+b)+c(b+c)(a+c)}{(b+c)(a+c)(a+b)} = 1. $$ But I don't have some success ahead.

Answer 1

By assumption we have $$ \frac{a^3 + abc + b^3 + c^3}{(a + b)(a + c)(b + c)}=0. $$ The second term equals $$ \dfrac{a^2}{b+c} + \dfrac{b^2}{a+c} + \dfrac{c^2}{a+b} =\frac{(a^3 + abc + b^3 + c^3)(a+b+c)}{(a + b)(a + c)(b + c)} $$ So it is equal to zero.