If $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1$, what can we say about $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}$? 
Suppose that $a,b,c$ are three real numbers such that $\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1$.  What are the possible values for $\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}$?

After clearing the denominators, we have $$a(c+a)(a+b)+b(b+c)(a+b)+c(b+c)(c+a)=(a+b)(b+c)(a+c)\,.$$
That is,
$$a^3+b^3+c^3+abc=0\,.$$
But then I'm stuck.  This question is related, but a bit different.
Thank you for your help!

Hint. You have
$$ \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} - 1 = \frac{a^3+b^3+c^3+abc}{(a+b)(b+c)(c+a)} $$
as well as
$$ \frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = \frac{(a^3+b^3+c^3+abc)(a+b+c)}{(a+b)(b+c)(c+a)}. $$
