Unable to reconstruct a new polynomial to find a given value Suppose $f(x)= x^3+2x^2+3x+3$ and has roots $a , b ,c$.
Then find the value of
$\left(\frac{a}{a+1}\right)^{3}+\left(\frac{b}{b+1}\right)^{3}+\left(\frac{c}{c+1}\right)^{3}$.
My Approach :
I constructed a new polynomial $g(x) = f\left(\frac{x^{\frac{1}{3}}}{1-x^{\frac{1}{3}}}\right)$ and then used the Vieta's formula for the sum of roots taken one at a time to solve the sum.
But then I realised that I won't be able to do so as $g(x)$ is not a polynomial anymore.
Can anyone help me please.
 A: It has some boring calculations so I'll just write about the sketch of my solution:
Using Vietta formulas, find the coefficients of the polynomial that has roots ${1-\frac{1}{a+1},1-\frac{1}{b+1},1-\frac{1}{c+1}}$. Then just use the Newton formula to find the sum of cubes of that polynomial
A: We shall make use of the following well-known identity:
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz).$$
Now, using Vieta's Formula, $a+b+c=-2, ab+ac=bc=3$, and $abc=-3$.
Thus, by direct expansion, we have that
$(a+1)(b+1)(c+1)=-1.$
\begin{align} 
\dfrac{a}{a+1}+\dfrac{b}{b+1} + \dfrac{c}{c+1} &= 3- \left(\dfrac{1}{a+1} + 
\dfrac{1}{b+1} + \dfrac{1}{c+1} \right) \\
& = 3- \dfrac{(b+1)(c+1) + (a+1)(c+1) + (b+1)(a+1)}{-1} \\
& = 3   + bc+b+c+1 + ac+ a + c+1  + ab+ a +b+1 \\
& = 5.
\end{align}
$$\dfrac{a}{a+1} \cdot \dfrac{b}{b+1} \cdot \dfrac{c}{c+1} = \dfrac{abc}{-1}=3.$$
\begin{align}
\dfrac{a}{a+1}\cdot \dfrac{b}{b+1} + \dfrac{a}{a+1}\cdot \dfrac{c}{c+1} + \dfrac{b}{b+1}\cdot \dfrac{c}{c+1} & = \dfrac{3(c+1)}{c} + \dfrac{3(b+1)}{b} + \dfrac{3(a+1)}{a} \\
& = 3 \left(3+ \dfrac{1}{c} + \dfrac{1}{b} + \dfrac{1}{a} \right) \\
& = 3 \left(3 + \dfrac{ab+bc+ca}{abc}\right) \\
& = 6.
\end{align}
Let $A=\dfrac{a}{a+1}, B=\dfrac{b}{b+1} , C=\dfrac{c}{c+1}$. Then,
\begin{align}
A^3 + B^3 + C^3 & = (A+B+C)(A^2+B^2+C^2-AB-BC-CA)+3ABC \\
& = 5 \left((A+B+C)^2-2(AB+BC+CA)-AB-BC-CA \right) + 3ABC& \\
& = 5 (25 -3(6))+3(3) \\
& = \mathbf{44}.
\end{align}
