Question

$\alpha$ &$\beta$ are the roots of the equation $x^2-7x-1=0$, then $\frac{{\alpha }^{10}+{\beta }^{10}-({\alpha }^8+{\beta }^8)}{{\alpha }^{\alpha +\beta +2}-{\beta }^{\alpha +\beta +2}}$.

Answer 1

Since $\alpha ,\beta$ are roots

So, ${\alpha }^2-7\alpha -1=0$

and ${\beta }^2-7\beta -1=0$

for ${\alpha }^2-7\alpha -1=0$

Product of roots$=-1$

So, roots are $\alpha ,\frac{-1}{\alpha }$

So, $\alpha +\frac{1}{\alpha }=\alpha -\left(\frac{-1}{\alpha }\right)=\sqrt{{\left(7\right)}^2-4(-1)}=\sqrt{53}$

Similarly, $\beta +\frac{1}{\beta }=\sqrt{53}$

and $\alpha -\frac{1}{\alpha }=\beta -\frac{1}{\beta }=7$

Now, $\alpha +\beta =7$

$2+\alpha +\beta =7+2=9$

So, $\frac{{\alpha }^{10}+{\beta }^{10}-({\alpha }^8+{\beta }^8)}{{\alpha }^{\alpha +\beta +2}-{\beta }^{\alpha +\beta +2}}=\frac{{\alpha }^9(\alpha -\frac{1}{\alpha })-{\beta }^9(\beta -\frac{1}{\beta })}{{\alpha }^9-{\beta }^9}$

$=\frac{{\alpha }^9\times 7-{\beta }^9\times 7}{{\alpha }^9-{\beta }^9}=\frac{7({\alpha }^9-{\beta }^9)}{{\alpha }^9-{\beta }^9}=7$