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


For the first, as Avatar described, you use that $\frac{1}{\alpha+1}$ satisfies the polynomial relation $6y^3-7y^2+3y-1=0$ to express its third power in lower powers of itself.
For the second, you rewrite it simply as $\frac{(\alpha+1)(\beta+1)(\gamma+1)}{(\alpha+1)^3}+\frac{(\alpha+1)(\beta+1)(\gamma+1)}{(\beta+1)^3}+\frac{(\alpha+1)(\beta+1)(\gamma+1)}{(\gamma+1)^3}$ which reduces the problem to calculating $(\alpha+1)(\beta+1)(\gamma+1)$, which being a symmetric rational function in the roots should be easy to relate to the coefficients of $6y^3-7y^2+3y-1$.
