$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 9$ Let $$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 9,$$
$$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = 32,$$
$$\frac{a^3}{b+c} + \frac{b^3}{c+a} + \frac{c^3}{a+b} = 122.$$
Find the value of $abc$.
Please check if my answer is correct or not.
$a+b+c + \frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = 32 +a+b+c$
$(a+b+c)\left(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}\right) = 32 +a+b+c$
$a+b+c = 4$
$a^2+b^2+c^2 = 6$
$ab+bc+ca = 5$
$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} + 3= 12$
$(a+b+c)\left(\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b}\right) = 12$
$\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} = 3$
$\frac{a^2+b^2+c^2+3(ab+bc+ca)}{2abc+\sum{a^2b} + \sum{ab^2}} = 3$
$7 = (ab+bc+ca)(a+b+c)-abc$
$abc = 13$

Your solution is correct. 
You find $a^2+b^2+c^2=6$ using the same trick:
$$\frac{a^3}{b+c} + \frac{b^3}{c+a} + \frac{c^3}{a+b} = 122 \iff \\
(a+b+c)\left(\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b}\right)-(a^2+b^2+c^2) = 122\Rightarrow \\
a^2+b^2+c^2=4\cdot 32-122=6.$$
Then:
$$ab+bc+ca=\frac12[(a+b+c)^2-(a^2+b^2+c^2)]=\frac12[4^2-6]=5.$$
For the following steps useful formulas:
$$\begin{align}(a+b+c)^3&=a^3+b^3+c^3+3(a+b+c)(ab+bc+ca)-3abc \ \ (1)\\
(a+b+c)^3&=a^3+b^3+c^3+3(a+b)(b+c)(c+a) \ \ (2)\end{align}$$
From the first equation:
$$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 9\Rightarrow \\
9(a+b)(b+c)(c+a)=a^3+b^3+c^3+(a+b+c)(ab+bc+ca)$$
Considering this with $(2)$:
$$3(a+b+c)^3-3(a^3+b^3+c^3)=a^3+b^3+c^3+(a+b+c)(ab+bc+ca) \Rightarrow \\
a^3+b^3+c^3=3\cdot 4^3-4\cdot 5=43.$$
From $(1)$:
$$abc=\frac{43+3\cdot 4\cdot 5-4^3}{3}=13.$$
