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 $(a+b+c)^3=a^3+b^3+c^3-3(a+b)(b+c)(c+a)$
Best Answer
Substituting $x=\frac y{1-y}$, which is the inverse of $y=\frac x{x+1}$, yields $$y^3-5y^2+6y-3=0$$ and $a,b,c=\frac\alpha{1+\alpha},\frac\beta{1+\beta},\frac\gamma{1+\gamma}$ are its roots. Then the desired expression is $$a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+bc+ca)+3abc$$ and we can extract the values of the symmetric polynomials from the coefficients: $a+b+c=5,ab+bc+ca=6,abc=3$ and $$a^3+b^3+c^3=5^3-3\cdot5\cdot6+3\cdot3=44$$