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
Best Answer
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.