$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 9$ 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.
$a+b+c + \frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = 32 +a+b+c$
$(a+b+c)\left(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}\right) = 32 +a+b+c$
$a+b+c = 4$
$a^2+b^2+c^2 = 6$
$ab+bc+ca = 5$
$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} + 3= 12$
$(a+b+c)\left(\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b}\right) = 12$
$\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} = 3$
$\frac{a^2+b^2+c^2+3(ab+bc+ca)}{2abc+\sum{a^2b} + \sum{ab^2}} = 3$
$7 = (ab+bc+ca)(a+b+c)-abc$
$abc = 13$
 A: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=9$, eqI
$\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=32$, eqII
$\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}=122$ eqIII
Let $a+b+c=k \Rightarrow$ 
$ a=k-(b+c), b=k-(a+c), c=k-(a+b)$. (eqIV)
Using (eqI): $\frac{k-(b+c)}{b+c}+\frac{k-(a+c)}{a+c}+\frac{k-(a+b)}{a+b}=9 \Rightarrow \frac{k}{b+c}-1+\frac{k}{a+c}-1+\frac{k}{a+b}-1=9 \Rightarrow$
$ k(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})=12 $ (eqV).
Lets apply (eqIV) in (eqII): 
$\frac{(k-(b+c))²}{b+c}+\frac{(k-(a+c))²}{a+c}+\frac{(k-(a+b))²}{a+b}=32$
$\frac{k²-2k(b+c)+(b+c)²}{b+c}+\frac{k²-2k(a+c)+(a+c)²}{a+c}+\frac{k²-2k(a+b)+(a+b)²}{a+b}=32$
$k²(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})-6k+(b+c)+(a+c)+(a+b)=32$
$k.[k(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})]-4k=32$
Using (eqV):
$ k.12-4k=32 \Rightarrow k=4$
Lets do the same with (eqIV) in (eqIII):
$\frac{(k-(b+c))³}{b+c}+\frac{(k-(a+c))³}{a+c}+\frac{(k-(a+b))³}{a+b}=122$
$\frac{k³-3k²(b+c)+3k(b+c)²+(b+c)³}{b+c}+\frac{k³-3k²(a+c)+3k(a+c)²+(a+c)³}{a+c}+\frac{k³-3k²(a+b)+3k(a+b)²+(a+b)³}{a+b}=122$
$k²(\frac{k}{b+c}+\frac{k}{a+c}+\frac{k}{a+b})
-3k²(1+1+1)+3k((b+c)+(a+c)+(a+b))-((b+c)²+(a+c)²+(a+b)²)=122$
$k²(12)-9k²+3k(2k)-2(a²+b²+c²+ab+bc+ac)=122$
Using that k=4, $144-2(a²+b²+c²+ab+bc+ac)=122$
$ a²+b²+c²+ab+bc+ac = 11$
But we know that $k=a+b+c=4 \Rightarrow (a+b+c)²=(a²+b²+c²)+2(ab+bc+ac)$
Lets call $S_2=a²+b²+c²$ and $\sigma_2=ab+bc+cd$. So
$ S_2+\sigma_2= 11$
$ S_2+2\sigma_2= 16$
Solving this, we have $S_2=6$ and $\sigma_2=5$.
But $(a+b+c)^3=a³+b³+c³+3(a+b+c)(\sigma_2)-3abc \Rightarrow S_3=a³+b³+c³=4^3-3*4*5+3abc$
$\Rightarrow S_3=4+3abc$
$ k(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})=12 $
$ (\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})=3 $
$ \frac{(b+c)(a+c)+(b+c)(a+b)+(a+c)(a+b)}{(b+c)(a+c)(a+b)}= 3$
$ \frac{(ab+ac+bc+c²)+(ba+b²+ac+bc)+(a²+ab+ac+bc)}{a²(b+c)+b²(a+c)+c²(a+b)+2abc}=3 $
$ \frac{S_2+3\sigma_2}{a²(4-a)+b²(4-b)+c²(4-c)+2abc}=3 $
$ \frac{S_2+3\sigma_2}{4S_2-S_3+2abc}=3 $
$ \frac{6+3*5}{4*6-(4+3abc)+2abc}=3 $
$ abc=13 $
A: 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 $(1)$:
$$abc=\frac{43+3\cdot 4\cdot 5-4^3}{3}=13.$$
