<s>[INST] Manipulating Algebraic Expression 
$a + b + c = 7$ and $\dfrac{1}{a+b} + \dfrac{1}{b+c} + \dfrac{1}{c+a} = \dfrac{7}{10}$. Find the value of $\dfrac{a}{b+c} + \dfrac{b}{c+a} + \dfrac{c}{a+b}$.

I algebraically manipulated the second equation to get:
$\dfrac{(b+c)(c+a) + (a+b)(c+a) + (a+b)(b+c)}{(a+b)(b+c)(c+a)} = \dfrac{7}{10}$ 
$\dfrac{bc+ab+c^2+ac+a^2+bc+ba+ab+ac+b^2+bc}{(a+b)(b+c)(c+a)} = \dfrac{7}{10}$
$\dfrac{(a+b+c)^2}{(a+b)(b+c)(c+a)} = \dfrac{7}{10}$
$\dfrac{7^2}{(a+b)(b+c)(c+a)} = \dfrac{7}{10}$
$(a+b)(b+c)(c+a) = 70$
I'm stuck after this. 
 [/INST] HINT:
$$\frac a{b+c}+\frac b{c+a}+\frac c{a+b}$$
$$=\frac a{b+c}+1-1+\frac b{c+a}+1-1+\frac c{a+b}+1-1$$
$$=-3+(a+b+c)\left(\frac 1{b+c}+\frac 1{c+a}+\frac 1{a+b}\right)$$
Using summation notation, $$\sum_{a,b,c} \frac a{b+c}=-3+\sum_{a,b,c} \left(\frac a{b+c}+1\right)=-3+(a+b+c)\sum_{a,b,c}\frac1{b+c}$$
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