Question

A=√8-√7 and a=1/b then fine a^2+b^2-3ab/a^2+ab+b^2

Answer 1

$\dfrac{a^2+b^2-3ab}{a^2+ab+b^2}=\dfrac{27}{31}$

Step-by-step explanation:

We have,

$a=\sqrt{8} -\sqrt{7}$ and $a=\dfrac{1}{b}$

To find, $\dfrac{a^2+b^2-3ab}{a^2+ab+b^2}=?$

∴$b=\dfrac{1}{a}=\dfrac{1}{\sqrt{8} -\sqrt{7}}$

Rationalising numerator and denominator, we get

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

$b=\dfrac{\sqrt{8} +\sqrt{7}}{\sqrt{8}^2 -\sqrt{7}^2}=\dfrac{\sqrt{8} +\sqrt{7}}{8 -7}$

Using algebraic identity,

$a^{2} -b^{2} =(a+b)(a-b)$]

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

∴ $\dfrac{a^2+b^2-3ab}{a^2+ab+b^2}$

= $\dfrac{(a-b)^2-ab}{(a+b)^2-ab}$

$=\dfrac{[(\sqrt{8} -\sqrt{7})-(\sqrt{8} +\sqrt{7})]^2-(\sqrt{8} -\sqrt{7})(\sqrt{8} +\sqrt{7})}{[(\sqrt{8} -\sqrt{7})-(\sqrt{8} +\sqrt{7})]^2-(\sqrt{8}-\sqrt{7})(\sqrt{8} +\sqrt{7})}$

$=\dfrac{(\sqrt{8} -\sqrt{7}-\sqrt{8} -\sqrt{7})^2-(\sqrt{8}^2 -\sqrt{7}^2)}{(\sqrt{8} -\sqrt{7}+\sqrt{8} +\sqrt{7})^2-(\sqrt{8}^2- \sqrt{7}^2)}$

$=\dfrac{(-2\sqrt{7})^2-1}{(2\sqrt{8})^2-1}$

$=\dfrac{28-1}{32-1}$

$=\dfrac{27}{31}$

∴ $\dfrac{a^2+b^2-3ab}{a^2+ab+b^2}=\dfrac{27}{31}$