<s>[INST] Find the value of $(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3$ Find the value of

$(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3$ where $\alpha,\beta,\gamma$ are roots of the equation $x^3+2x^2+3x+3=0$.

I tried to use the formula which is wrong

$(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3=((\frac{\alpha}{\alpha +1})+(\frac{\beta}{\beta +1})+(\frac{\gamma}{\gamma +1}))^3-3((\frac{\alpha}{\alpha +1})(\frac{\beta}{\beta +1})(\frac{\gamma}{\gamma +1}))$

And then I broke the terms to get

$(\frac{13}{5})^3-\frac{36}{5}$

but this is $\ne 44$, which should be the answer.
I know where I was wrong, the formula is $(a+b+c)^3=a^3+b^3+c^3-3(a+b)(b+c)(c+a)$
 [/INST] I think your way is not really practical. Here it is a shortcut:
$$ \begin{array}{|c|c|}\hline \text{Roots}&\text{Polynomial}\\
\hline \alpha,\beta,\gamma & x^3+2x^2+3x+3\\
\hline \frac{1}{\alpha},\frac{1}{\beta},\frac{1}{\gamma} & x^3+x^2+\frac{2}{3}x+\frac{1}{3}\\
\hline 1+\frac{1}{\alpha},1+\frac{1}{\beta},1+\frac{1}{\gamma} & x^3-2x^2+\frac{5}{3}x-\frac{1}{3}\\
\hline \frac{\alpha}{\alpha+1},\frac{\beta}{\beta+1},\frac{\gamma}{\gamma+1}&x^3-5x^2+6x-3 \\ \hline
 \end{array} $$
Now by setting $A=\frac{\alpha}{\alpha+1},B=\frac{\beta}{\beta+1},C=\frac{\gamma}{\gamma+1}$ we have $A+B+C=5$ and $AB+AC+BC=6$ by Vieta's formulas, hence $A^2+B^2+C^2=25-12=13$. Additionally for any $z\in\{A,B,C\}$ we have $z^3=5z^2-6z+3$, hence
$$ A^3+B^3+C^3 = 5\cdot 13-6\cdot 5+3\cdot 3 = 65-30+9 = \color{red}{44}.$$
An equivalent way is to compute the trace of $\left(M(M+I)^{-1}\right)^3=(I+M^{-1})^{-3}$, where $M$ is the companion matrix of $x^3+2x^2+3x+3$.
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