Calculate $\frac{2}{\alpha^2}-\frac{1}{(\alpha +1)^2}$ if $\alpha$ be a root of $x^2+(1-\sqrt3)x+1-\sqrt3=0$

Question

If $\alpha$ is a root of $x^2+(1-\sqrt3)x+1-\sqrt3=0$ calculate

$$\frac{2}{\alpha^2}-\frac{1}{(\alpha +1)^2}$$ What I have done: $$\begin{aligned} \alpha^2+\alpha+1&=\sqrt3(\alpha+1)\\ &\implies\alpha^4+2\alpha^3+3\alpha^2+2\alpha+1=3\alpha^2+6\alpha+3\\ &\implies\alpha^2(\alpha+1)^2=\alpha^2+4\alpha+2\\ &\implies\frac{2}{\alpha^2}-\frac{1}{(\alpha +1)^2}=1 \end{aligned} $$

Is there any easy way to calculate?

Answer 1

Another way (not necessarily easier than OP’s one).

$\sqrt3(\alpha+1)=\alpha(\alpha+1)+1$

$\dfrac{\sqrt3}{\alpha}=1+\dfrac1{\alpha(\alpha+1)}$

$\left(\dfrac{\sqrt3}{\alpha}\right)^2=\left(1+\dfrac1\alpha-\dfrac1{\alpha+1}\right)^2$

$\dfrac3{\alpha^2}=1+\dfrac1{\alpha^2}+\dfrac1{(\alpha+1)^2}\;\;$ (by simplifying twice products)

$\dfrac2{\alpha^2}-\dfrac1{(\alpha+1)^2}=1$