Question

If a = √8 - √7 and a = 1/b, then (a^2 + b^2 - 3ab)/(a^2 + ab + b^2) is equal to:​

Answer 1

GIVEN :–

$\\ \: \: { \huge{.}} \: \: { \bold{a = \sqrt{8} - \sqrt{7} }}$

$\\ \: \: { \huge{.}} \: \: { \bold{a = \dfrac{1}{b} }}$

TO FIND :–

$\\ \: \: { \huge{.}} \: \: { \bold{ \frac{ {a}^{2} + {b}^{2} - 3ab }{ {a}^{2} + {b}^{2} + ab } = ? }}$

SOLUTION :–

• According to the question –

$\\ \: \implies \: \: { \bold{ a = \dfrac{1}{b} }}$

$\\ \: \implies \: \: { \bold{ b = \dfrac{1}{a} }}$

$\\ \: \implies \: \: { \bold{ b = \dfrac{1}{ \sqrt{8} - \sqrt{7} } }}$

• Now rationalization –

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

$\\ \: \implies \: \: { \bold{ b = \dfrac{ \sqrt{8} + \sqrt{7} }{ (\sqrt{8} + \sqrt{7} )(\sqrt{8} - \sqrt{7})} }}$

$\\ \: \implies \: \: { \bold{ b = \dfrac{ \sqrt{8} + \sqrt{7} }{ {( \sqrt{8} )}^{2} - {( \sqrt{7}) }^{2} } \: \: [ \: \: \because \: \: (a + b)(a - b) = {a}^{2} - {b}^{2} ]}}$

$\\ \: \implies \: \: { \bold{ b = \dfrac{ \sqrt{8} + \sqrt{7} }{ 8 - 7 }}}$

$\\ \: \implies \: \: { \bold{ b = { \sqrt{8} + \sqrt{7} }}}$

• Now –

$\\ \: \: = \: \: { \bold{ \frac{ {a}^{2} + {b}^{2} - 3ab }{ {a}^{2} + {b}^{2} + ab } }}$

$\\ \: \: = \: \: { \bold{ \frac{ {a}^{2} + {b}^{2} - 2ab - ab }{ {a}^{2} + {b}^{2} + 2ab - ab } }}$

$\\ \: \: = \: \: { \bold{ \frac{ {(a - b)}^{2} - ab }{ {(a + b)}^{2} - ab } \: \: \: \: \: \: [ \: \: \because \: \: {(a \pm b)}^{2} = {a}^{2} + {b}^{2} \pm 2ab \: ]}}$

• Now put the values –

$\\ \: \: = \: \: { \bold{ \frac{ {( \sqrt{8} - \sqrt{7} - \sqrt{8} - \sqrt{7} )}^{2} - ( \dfrac{1}{ \cancel b}) \cancel b }{ {( \sqrt{8} - \sqrt{7} + \sqrt{8} + \sqrt{7} )}^{2} - ( \dfrac{1}{ \cancel b}) \cancel b } }}$

$\\ \: \: = \: \: { \bold{ \frac{ {( - 2 \sqrt{7} )}^{2} - 1}{ {(2 \sqrt{8} ) }^{2} - 1} }}$

$\\ \: \: = \: \: { \bold{ \frac{ 28 - 1}{ 32- 1} }}$

$\\ \: \: = \: \: { \bold{ \frac{ 27}{ 31} }}$

• Hence , $\: \: { \bold{ \frac{ {a}^{2} + {b}^{2} - 3ab }{ {a}^{2} + {b}^{2} + ab } = \dfrac{27}{31} }}$

$\\ \rule{220}{2}$