Question

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Find the value of (a⁴ + b⁴) if $a = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3}- \sqrt{2} } \: and \\ b = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3}+ \sqrt{2} }$

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Answer 1

The answer is given below :

Now,
$a = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ = \frac{ {( \sqrt{3} + \sqrt{2}) }^{2} }{3 -2 } \\ \\ = \frac{3 + 2 \sqrt{6} + 2}{1} \\ \\ = 5 + 2 \sqrt{6}$

Again,

$b = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ = \frac{ {( \sqrt{3} - \sqrt{2}) }^{2} }{3 -2 } \\ \\ = \frac{3 - 2 \sqrt{6} + 2}{1} \\ \\ = 5 - 2 \sqrt{6}$

So,

${a}^{2} = {(5 + 2\sqrt{6} )}^{2} \\ \\ = 25 + 20 \sqrt{6} + 24 \\ \\ = 49 + 20 \sqrt{6} \\ \\ and \\ \\ {b}^{2} = {(5 - 2 \sqrt{6} )}^{2} \\ \\ = 25 - 20 \sqrt{6} + 24 \\ \\ = 49 - 20 \sqrt{6}$

Now,

${a}^{2} + {b}^{2} \\ \\ = (49 + 20 \sqrt{6} ) + (49 - 20 \sqrt{6} ) \\ \\ = 98 \\ \\ and \\ \\ {a}^{2} {b}^{2} \\ \\ = (49 + 20 \sqrt{6} )(49 - 20 \sqrt{6} ) \\ \\ = {49}^{2} - {(20 \sqrt{6} )}^{2} \\ \\ = 2401 - 2400 \\ \\ = 1$

Therefore,

${a}^{4} + {b}^{4} \\ \\ = {( {a}^{2} + {b}^{2} ) }^{2} - 2 {a}^{2} {b}^{2} \\ \\ = {98}^{2} - (2 \times 1) \\ \\ = 9604 - 2 \\ \\ = 9602$

Thank you for your question.

Answer 2

hy
here is your answer bro
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i dont know how to write on the formula box thats why i solved in copy