Value of expression in the roots of the cubic obtained by shifting $x^3-3x^2+x+2$ to eliminate the $x^2$ term 
For a given function $f(x)=x^3-3x^2+x+2$, roots are shifted by $\lambda$ to eliminate $x^2$ terms and $\alpha$, $\beta$, $\gamma$ be the roots of new function. Evaluate
$$\begin{align}
\frac{\alpha^3}{(\alpha-\beta)(\alpha-\gamma)(\alpha-\lambda)}&+\frac{\beta^3}{(\beta-\gamma)(\beta-\alpha)(\beta-\lambda)}+\frac{\gamma^3}{(\gamma-\alpha)(\gamma-\beta)(\gamma-\lambda)} \\
&+\frac{\lambda^3}{(\lambda-\alpha)(\lambda-\beta)(\lambda-\gamma)}
\end{align}$$

What I try
Replace $x\rightarrow x+\lambda$ in $f(x)=x^3-3x^2+x+2$
$f(x+\lambda) = (x+\lambda)^3-3(x+\lambda)^2+(x+\lambda)+2$
$x^3+\lambda^3+3x^2\lambda+3x\lambda^2-3x^2-3\lambda^2-6x\lambda+x+\lambda+2$
$x^3+(3\lambda-3)x^2+(3\lambda^2-6\lambda+1)x+\lambda^3-3\lambda^2+\lambda+2=0$
put $3\lambda-3=0\Rightarrow \lambda = 1$
So equation is $x^3-2x+1=0$ has roots $\alpha,\beta,\gamma$
So we have $\alpha+\beta+\gamma=0,\sum \alpha \beta=-2,\alpha\beta\gamma=-1$
How do I solve it Help me please

Hint: $(x^3-2x+1)$=$(x-1)(x^2+x-1)$
