<s>[INST] Relations between the roots of a cubic polynomial How do I solve the last two of these problems?

The roots of the equation $x^3+4x-1=0$ are $\alpha$, $\beta$, and $\gamma$. Use the substitution $y=\dfrac{1}{1+x}$ to show that the equation $6y^3-7y^2+3y-1=0$ has roots $\dfrac{1}{\alpha+1}$, $\dfrac{1}{\beta+1}$, and $\dfrac{1}{\gamma+1} $. $\quad \quad \quad (2)$  
For the cases $n=1$ and $n=2$, find the value of
  $$\dfrac{1}{(\alpha+1)^n}+\dfrac{1}{(\beta+1)^n}+\dfrac{1}{(\gamma+1)^n}. \tag{2}$$
  Deduce the value of $\dfrac{1}{(\alpha+1)^3}+\dfrac{1}{(\beta+1)^3}+\dfrac{1}{(\gamma+1)^3}. \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \, \, \,(2)$  
Hence show that $\dfrac{(\beta+1)(\gamma+1)}{(\alpha+1)^2}+\dfrac{(\gamma+1)(\alpha+1)}{(\beta+1)^2}+\dfrac{(\alpha+1)(\beta+1)}{(\gamma+1)^2}=\dfrac{73}{36} \quad \quad \quad \quad \quad \quad \quad (3)$  

 [/INST] Part 1,
Since , $\frac{1}{\alpha +1}$ is the root of the equation $6y^3-7y^2+3y-1=0\implies $ $$\frac{1}{(\alpha +1)^3}=\frac{7}{(\alpha +1)^2}-\frac{3}{\alpha +1}+1$$
Similar equality follows for $\beta, \gamma$
Part 2,
Product of roots= $\frac{1}{6}\implies \frac{1}{1+\alpha}\frac{1}{1+\beta}\frac{1}{1+\gamma}=\frac{1}{6}\implies (1+\beta)(1+\gamma)=\frac{6}{1+\alpha}\implies \frac{(1+\beta)(1+\gamma)}{(1+\alpha)^2}=\frac{6}{(1+\alpha)^3}$
Therefore, $\frac{(1+\beta)(1+\gamma)}{(1+\alpha)^2}+\frac{(1+\beta)(1+\gamma)}{(1+\alpha)^2}+\frac{(1+\beta)(1+\gamma)}{(1+\alpha)^2}=6$(result of second part)
Now, you can easily calculate the last sum.
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