Question

Solve $\frac{\sqrt{7}-1}{\sqrt{7}+1}-\frac{\sqrt{7}+1}{\sqrt{7}-1}=a+b\sqrt{7}$

Answer 1

We have,

$\frac{\sqrt{7}-1}{\sqrt{7}+1}-\frac{\sqrt{7}+1}{\sqrt{7}-1}=a+b\sqrt{7}$

$\frac{{(\sqrt{7}-1)}^2-{(\sqrt{7}+1)}^2}{{\left(\sqrt{7}\right)}^2-1^2}=a+b\sqrt{7}$

$\frac{7+1-2\sqrt{7}-(7+1+2\sqrt{7})}{7-1}=a+b\sqrt{7}$

$\frac{7+1-2\sqrt{7}-7-1-2\sqrt{7}}{6}=a+b\sqrt{7}$

$\frac{-4\sqrt{7}}{6}=a+b\sqrt{7}$

$a+b\sqrt{7}=0-\frac{2\sqrt{7}}{3}$

Hence, the value of $a$ = $0$ and $b$ = $-\frac{2}{3}$.