TITLE: Find the value of $(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3$
QUESTION [6 upvotes]: 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 $(a+b+c)^3=a^3+b^3+c^3-3(a+b)(b+c)(c+a)$

REPLY [3 votes]: For simplicity, let me use $\alpha=a, \beta=b,\gamma=c$. 
We can find:
$$\begin{cases}a+b+c=-2 \\ ab+bc+ca=3\\ abc=-3\end{cases} \Rightarrow \\
\begin{cases}\color{red}{a^2+b^2+c^2}=4-2(ab+bc+ca)=\color{red}{-2} \\ \color{blue}{a^2b^2+b^2c^2+c^2a^2}=9-2abc(a+b+c)=\color{blue}{-3}\end{cases}$$
We can express:
$$x^3+2x^2+3x+3=0 \Rightarrow (x+1)^3=x^3+3x^2+3x+1=x^2-2$$
Hence:
$$\left(\frac{a}{a +1}\right)^3+\left(\frac{b}{b +1}\right)^3+\left(\frac{c}{c +1}\right)^3=\\
\frac{a^3}{a^2-2}+\frac{b^3}{b^2-2}+\frac{c^3}{c^2-2}=\\
a+\frac{2a}{a^2-2}+b+\frac{2b}{b^2-2}+c+\frac{2c}{c^2-2}=\\
-2+2\cdot \frac{a(b^2-2)(c^2-2)+b(a^2-2)(c^2-2)+c(a^2-2)(b^2-2)}{(a^2-2)(b^2-2)(c^2-2)}=\\
-2+2\cdot \frac{abc(ab+bc+ca)-2(a^2b+ab^2)-2(a^2c+ac^2)-2(b^2c+bc^2)+4(a+b+c)}{a^2b^2c^2-2(\color{blue}{a^2b^2+b^2c^2+c^2a^2})+4(\color{red}{a^2+b^2+c^2})-8}=\\
-2+2\cdot \frac{-9-2ab(-2-c)-2ac(-2-b)-2bc(-2-a)-8}{9+6-8-8}=\\
-2+2\cdot \frac{-9+4(ab+bc+ca)+6abc-8}{-1}=\\
-2+2\cdot \frac{-9+12-18-8}{-1}=44.\\
$$