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. 

It is sometimes beneficial to start from the end
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=$$
$$\frac{7-b-c}{b+c}+\frac{7-a-c}{a+c}+\frac{7-a-b}{a+b}=$$
$$7\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)-3=$$
$$\frac{49}{10}-3=\frac{19}{10}$$
