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 $\dfrac{(\beta+1)(\gamma+1)}{(\alpha+1)^2}+\dfrac{(\gamma+1)(\alpha+1)}{(\beta+1)^2}+\dfrac{(\alpha+1)(\beta+1)}{(\gamma+1)^2}=\dfrac{73}{36} \quad \quad \quad \quad \quad \quad \quad (3)$  


Let $a=\frac1{(1+\alpha)}$ etc,
so, $a,b,c$ are the roots of $6t^3-7t^2+3t-1=0$
$\implies a+b+c=\frac 7 6, ab+bc+ca=\frac 3 6=\frac 12$ and $abc=\frac1 6$
So, $$a^3+b^3+c^3=a^3+b^3+c^3-3abc+3abc$$
$$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$$
$$=(a+b+c)((a+b+c)^2-3(ab+bc+ca))+3abc$$
$$=(\frac 7 6)((\frac 7 6)^2-3(\frac 12))+3\frac1 6=\frac{73}{216}$$
Now, $$\frac{(\beta+1)(\gamma+1)}{(\alpha+1)^2}=\frac{a^2}{bc}=\frac{a^3}{abc}=6a^3$$
So, $$\sum \frac{(\beta+1)(\gamma+1)}{(\alpha+1)^2}=6(a^3+b^3+c^3)=6\left(\frac{73}{216}\right)=\frac{73}{36}$$
For the generalization of the sums of Powers of Roots, one may look here for the statement, here for the proof.
