Manipulating roots of a cubic Given that $A,B,C$ are the roots of the equation $x^3-5x^2+x+1$, how do I find the value of $$\dfrac{A}{B+C}+\dfrac{B}{A+C}+\dfrac{C}{A+B}$$ I know the Vieta's formulas but I am not able to manipulate the above expression into something known. And taking the LCM doesn't help. Please help me out. Thank you.

By Vieta's, you know that $A+B+C = 5$. Therefore, we have 
$\dfrac{A}{B+C}+\dfrac{B}{A+C}+\dfrac{C}{A+B} = \dfrac{A}{5-A}+\dfrac{B}{5-B}+\dfrac{C}{5-C}$
Also, you perform the following manipulation: 
$x^3-5x^2+x+1 = 0$ 
$5x^2-x^3 = x+1$
$\dfrac{5-x}{x} = \dfrac{x+1}{x^3}$
$\dfrac{x}{5-x} = \dfrac{x^3}{x+1}$
$\dfrac{x}{5-x} = x^2-x+1+\dfrac{1}{x+1}$
Now take the sum of both sides over $x = A,B,C$ and apply Vieta's. 

Alternatively, if $x = A,B,C$ are the roots of the equation $f(x) = x^3-5x^2+x+1 = 0$, then $x = \dfrac{A}{5-A}, \dfrac{B}{5-B}, \dfrac{C}{5-C}$ are the roots of $f\left(\dfrac{5x}{x+1}\right) = 0$. Apply Vieta's to that equation.
