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
 A: Hint: $(x^3-2x+1)$=$(x-1)(x^2+x-1)$
A: (As bjorn93 commented, if we consider $f(x+\lambda)$, then we get $\lambda=1$ and $(\lambda-\alpha)(\lambda-\beta)(\lambda-\gamma)=0$. So, this answer considers $f(x-\lambda)$.)
Since we have
$$\begin{align}f(x-\lambda)&=(x-\lambda)^3-3(x-\lambda)^2+(x-\lambda)+2
\\\\&=x^3+(-3\lambda-3)x^2+(3\lambda^2+6\lambda+1)x-3\lambda^2-\lambda+2\end{align}$$
we get $$-3\lambda-3=0\implies \lambda=-1$$
Then, we can write
$$x^3-2x+1=(x-1)(x^2+x-1)=(x-\alpha)(x-\beta)(x-\gamma)$$
with
$$\alpha+\beta+\gamma=0$$
Here, we may assume that $$\alpha=1\qquad\text{and}\qquad \gamma=-\beta-1$$
We also have
$$(\lambda-\alpha)(\lambda-\beta)(\lambda-\gamma)=f(0)=2$$
Multiplying top and bottom of
$$\frac{\alpha^3}{(\alpha-\beta)(\alpha-\gamma)(\alpha-\lambda)}$$
by $(\lambda-\beta)(\lambda-\gamma)(\beta-\gamma)$ and using $\alpha^3=2\alpha-1,(\lambda-\alpha)(\lambda-\beta)(\lambda-\gamma)=2$ gives
$$\frac{\alpha^3}{(\alpha-\beta)(\alpha-\gamma)(\alpha-\lambda)}=\frac{(2\alpha-1)(\lambda-\beta)(\lambda-\gamma)(\beta-\gamma)}{2(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)}$$
Similarly, we get
$$\frac{\beta^3}{(\beta-\gamma)(\beta-\alpha)(\beta-\lambda)}=\frac{(2\beta-1)(\lambda-\alpha)(\lambda-\gamma)(\gamma-\alpha)}{2(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)}$$
and
$$\frac{\gamma^3}{(\gamma-\alpha)(\gamma-\beta)(\gamma-\lambda)}=\frac{(2\gamma-1)(\lambda-\alpha)(\lambda-\beta)(\alpha-\beta)}{2(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)}$$
with
$$\frac{\lambda^3}{(\lambda-\alpha)(\lambda-\beta)(\lambda-\gamma)}=-\frac 12$$
From these, we have
$$\begin{align}
\frac{\alpha^3}{(\alpha-\beta)(\alpha-\gamma)(\alpha-\lambda)}&+\frac{\beta^3}{(\beta-\gamma)(\beta-\alpha)(\beta-\lambda)}+\frac{\gamma^3}{(\gamma-\alpha)(\gamma-\beta)(\gamma-\lambda)} \\
&+\frac{\lambda^3}{(\lambda-\alpha)(\lambda-\beta)(\lambda-\gamma)}
\end{align}$$
$$=-\frac 12+\frac{A}{2(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)}=-\frac 12+\frac{A}{2(\beta-1)(2\beta+1)(\beta+2)}$$
where
$$A=(2\alpha-1)(\lambda-\beta)(\lambda-\gamma)(\beta-\gamma)+(2\beta-1)(\lambda-\alpha)(\lambda-\gamma)(\gamma-\alpha)+(2\gamma-1)(\lambda-\alpha)(\lambda-\beta)(\alpha-\beta)$$
$$=(-1-\beta)\beta(2\beta+1)-2(2\beta-1)\beta(-2-\beta)-2(-3-2\beta)(-1-\beta)(1-\beta)$$
$$=3 (2 \beta^3 + 3 \beta^2 - 3\beta - 2)=3  (\beta - 1)(2\beta + 1) (\beta + 2)$$
Therefore, the answer is
$$-\frac 12+\frac{3(\beta-1)(2\beta+1)(\beta+2)}{2(\beta-1)(2\beta+1)(\beta+2)}=-\frac 12+\frac 32=\color{red}{1}$$
