Question

if a=√5+1/√5-1 and b=√5-1/√5+1 find the value of (a-b) ^3 / (a+b) ^3​

Answer 1

Given :

→$a = \frac{ \sqrt{5} + 1}{ \sqrt{5} - 1}$

→$b = \frac{ \sqrt{5} - 1 }{ \sqrt{5} + 1}$

To Find :

$\frac{ {(a - b)}^{3} }{ {(a + b)}^{3} }$

Solution :

$a = \frac{ \sqrt{5} + 1 }{ \sqrt{5} - 1}$

Upon rationalising a, we get

$a = \frac{ \sqrt{5} + 1}{ \sqrt{5} - 1 } \times \frac{ \sqrt{5} + 1 }{ \sqrt{5} + 1}$

By using, (a+b) ² and (a²-b²) =(a+b) (a-b) we have

$a = \frac{ { \sqrt{5} }^{2} + {1}^{2} + 2 \sqrt{5} }{( { \sqrt{5}) }^{2} - {1}^{2} } \\ \\ a = \frac{5 + 1 + 2 \sqrt{5} }{5 - 1} \\ \\ a = \frac{6 + 2 \sqrt{5} }{4} = \frac{3 + \sqrt{5} }{2}$

And,

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$b = \frac{ \sqrt{5} - 1}{ \sqrt{5} + 1 }$

Upon rationalising b, we get

$b = \frac{ \sqrt{5} - 1}{ \sqrt{5} + 1} \times \frac{ \sqrt{5} - 1}{ \sqrt{5} - 1}$

By using, (a+b) ² and (a²-b²) = a²- b² we have

$b = \frac{ { \sqrt{5} }^{2} + {1}^{2} - 2 \sqrt{5} }{ { \sqrt{5} }^{2} - {1}^{2} } \\ \\ b = \frac{5 + 1 - 2 \sqrt{5} }{5 - 1} \\ \\ b = \frac{6 - 2 \sqrt{5} }{4} = \frac{3 - \sqrt{5} }{2}$

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Now, according to the question we need to find

$\frac{ {(a - b)}^{3} }{ {(a + b)}^{3} }$

$= > ( {\frac{\frac{3 + \sqrt{5} - 3 + \sqrt{5} )}{2} }{ \frac{3 + \sqrt{5} + 3 - \sqrt{5} }{2}})}^{3}$

$= > ({ \frac{ \frac{2 \sqrt{5} }{2} }{ \frac{6}{2} } }) ^{3}$

$= > ( {\frac{2 \sqrt{5} }{2} \times \frac{2}{6} }) ^{3}$

$= > ({ \frac{ \sqrt{5} }{3} })^ {3}$

$= > \frac{5 \sqrt{5} }{27}$