If  $$ a =\sqrt{8}-\sqrt{7} $$ and $$ a=\frac{1}{b} $$ then $$ \frac{a^{2}+b^{2}-3ab}{a^{2}+b^{2}+ab} $$  is equal to

Given, $$ a =\sqrt{8}-\sqrt{7} $$ and

$$ a=\frac{1}{b} $$ $$=$$> $$ b=\frac{1}{a} $$ $$=\frac{1}{\sqrt{8}-\sqrt{7}}$$

$$=$$> $$b=\frac{1}{\sqrt{8}-\sqrt{7}}\times\frac{\sqrt{8}+\sqrt{7}}{\sqrt{8}+\sqrt{7}}$$

$$=$$> $$b=\frac{\sqrt{8}+\sqrt{7}}{8-7}$$

$$=$$> $$b=\sqrt{8}+\sqrt{7}$$

$$\therefore\ $$ $$ \frac{a^{2}+b^{2}-3ab}{a^{2}+b^{2}+ab} =\frac{\left(a-b\right)^2-ab}{\left(a+b\right)^2-ab}=\frac{\left(-2\sqrt{7}\right)^2-\left[\left(\sqrt{8}-\sqrt{7}\right)\left(\sqrt{8}+\sqrt{7}\right)\right]}{\left(2\sqrt{8}\right)^2-\left[\left(\sqrt{8}-\sqrt{7}\right)\left(\sqrt{8}+\sqrt{7}\right)\right]}=\frac{4\left(7\right)-\left[8-7\right]}{4\left(8\right)-\left[8-7\right]}=\frac{28-1}{32-1}=\frac{27}{31}$$