$\alpha,\beta,\gamma$ are roots of cubic equation $x^3+4x-1=0$ If $\alpha,\beta,\gamma$ are the roots of the equation $x^3+4x-1=0$ and $\displaystyle \frac{1}{\alpha+1},\frac{1}{\beta+1},\frac{1}{\gamma+1}$ are the roots of the equation 
$\displaystyle 6x^3-7x^2+3x-1=0$. Then value of $\displaystyle \frac{(\beta+1)(\gamma+1)}{\alpha^2}+\frac{(\gamma+1)(\alpha+1)}{\beta^2}+\frac{(\alpha+1)(\beta+1)}{\gamma^2}=$
$\bf{My\; Try}::$ Given $x=\alpha,\beta,\gamma$ are the roots of the equation $x^3+4x-1=0$, Then 
$x^3+4x-1 = (x-\alpha)\cdot(x-\beta)\cdot (x-\gamma)$
put $x=-1$, we get $(1+\alpha)(1+\beta)(1+\gamma) = 6$
Now $\displaystyle \frac{(\beta+1)(\gamma+1)}{\alpha^2}+\frac{(\gamma+1)(\alpha+1)}{\beta^2}+\frac{(\alpha+1)(\beta+1)}{\gamma^2} = 6\left\{\frac{1}{(\alpha+1)\alpha^2}+\frac{1}{(\beta+1)\beta^2}+\frac{1}{(\gamma+1)\gamma^2}\right\}$
Now how can I solve after that,
Help required
thanks
 A: Your calculation leads us$$\displaystyle \frac{(\beta+1)(\gamma+1)}{\alpha^2}+\frac{(\gamma+1)(\alpha+1)}{\beta^2}+\frac{(\alpha+1)(\beta+1)}{\gamma^2} $$$$= 6\left\{\frac{1}{(\alpha+1)\alpha^2}+\frac{1}{(\beta+1)\beta^2}+\frac{1}{(\gamma+1)\gamma^2}\right\}.$$
Here, by Vieta's formulas, notice that
$$\frac{7}{6}=\frac{1}{\alpha+1}+\frac{1}{\beta+1}+\frac{1}{\gamma+1}=\frac{(\alpha\beta+\beta\gamma+\gamma\alpha)+2(\alpha+\beta+\gamma)+3}{(\alpha+1)(\beta+1)(\gamma+1)}.$$
Since $\alpha+\beta+\gamma=0,$ we have
$$(\alpha+1)(\beta+1)(\gamma+1)=6.$$
Noting that your last $\{\}$ equals to
$$\frac{({\alpha}^3+{\beta}^3+{\gamma}^3)+({\alpha}^2+{\beta}^2+{\gamma}^2)}{(\alpha\beta\gamma)^2(\alpha+1)(\beta+1)(\gamma+1)},$$ we can use
$${\alpha}^3+{\beta}^3+{\gamma}^3=3\alpha\beta\gamma+(\alpha+\beta+\gamma)({\alpha}^2+{\beta}^2+{\gamma}^2-\alpha\beta-\beta\gamma-\gamma\alpha).$$
with
$$\alpha+\beta+\gamma=0,\alpha\beta+\beta\gamma+\gamma\alpha=4,\alpha\beta\gamma=1$$
and
$${\alpha}^2+{\beta}^2+{\gamma}^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha).$$
Now you'll be able to get the answer.
A: Hint: Since $x^{3} + 4x - 1 = (x-\alpha)(x-\beta)(x-\gamma)$, we note that, by the theory of symmetric polynomials, 
$$0 = \alpha+\beta+\gamma = s_{1}(\alpha, \beta, \gamma)$$
$$4 = \alpha\beta + \beta\gamma+\alpha\gamma = s_{2}(\alpha, \beta, \gamma)$$
$$1 = \alpha\beta\gamma = s_{3}(\alpha, \beta, \gamma)$$
As suggested by Mariano Suárez-Alvarez, we can observe immediately that the expression
$\displaystyle \frac{(\beta+1)(\gamma+1)}{\alpha^2}+\frac{(\gamma+1)(\alpha+1)}{\beta^2}+\frac{(\alpha+1)(\beta+1)}{\gamma^2}$
is symmetric in $\alpha, \beta, \gamma$, i.e. it is invariant under permuting $\alpha, \beta, \gamma$. Therefore, I would put everything under a common denominator. The numerator then becomes
$$\beta^{2}\gamma^{2}(\beta\gamma+\gamma+\beta+1) + \alpha^{2}\gamma^{2}(\alpha\gamma+\gamma+\alpha+1) + \alpha^{2}\beta^{2}(\alpha\beta + \alpha + \beta+1)$$
It can easily be seen that this is a symmetric polynomial in $\alpha, \beta, \gamma$ as well. It can therefore be written in terms of $s_{1}, s_{2}, s_{3}$. The denominator is $\alpha^{2}\beta^{2}\gamma^{2} = s_{3}^{2} = 1^{2} = 1$. If you would like more information on how to write the numerator in terms of the symmetric polynomials, please comment and I'd be happy to add information.  
A: From the second equation, $$\frac6{(\alpha+1)^3}-\frac7{(\alpha+1)^2}+\frac3{\alpha+1}-1=0$$
$$\implies 6-7(\alpha+1)+3(\alpha+1)^2-(\alpha+1)^3=0\iff \alpha^3+4\alpha-1=0$$
So, both equations are actually equivalent
$\displaystyle y=\frac{(\beta+1)(\gamma+1)}{\alpha^2}=\frac{\beta\gamma+\beta+\gamma+1}{\alpha^2}=\frac{\alpha\beta\gamma+\alpha(\beta+\gamma)+\alpha}{\alpha^3}=\frac{1-\alpha^2+\alpha}{1-4\alpha} $ as  $\alpha^3=1-4\alpha$
$\displaystyle\implies \alpha^2-\alpha(1+4y)+y-1=0\  \ \  \ (1)$
$\displaystyle y=\frac6{\alpha^2(\alpha+1)}=\frac6{(1-4\alpha)+\alpha^2}$
$\displaystyle\implies \alpha^2y-4y\alpha+y-6=0\ \  \ \ (2) $
Solve $(1),(2)$ for  $\alpha^2,\alpha$ 
and use $\alpha^2=(\alpha)^2$ to eliminate $\alpha$ and form a cubic equation in $y$ whose roots are $\displaystyle y_1,y_2,y_3$(say)
Now use Vieta's formulas to find the required $\displaystyle y_1+y_2+y_3$
We can also find $\displaystyle y_1y_2+y_2y_3+y_3y_1$ from here if need be
