Question

Rationalise the denominator x=3-2root2 find x square +1/ x square , x cube +1/x cube

Answer 1

Hi Mate !!


x = 3 - 2√2


$\frac{1}{x} = \frac{1}{3 - 2 \sqrt{2} }$

$\frac{1}{x} = \frac{1}{3 - 2 \sqrt{2} } \times \frac{3 + 2 \sqrt{2} }{3 + 2 \sqrt{2} }$


$\frac{1}{x} = \frac{3 + 2 \sqrt{2} }{9 - 8} = \frac{3 + 2 \sqrt{2} }{1}$


[ Using identity :- ( a + b ) ( a - b ) = a² - b² in denominator ]


$\frac{1}{x} = \: \: 3 + 2 \sqrt{2}$


Now ,

$( {x + \frac{1}{x}) }^{2} = {x}^{2} + \frac{1}{ {x}^{2} } + 2x \times \frac{1}{x}$


[ Using identity : ( a + b )² = a² + b² + 2ab ]


$( {3 - 2 \sqrt{2} + 3 + 2 \sqrt{2} \: \: \: ) }^{2} = {x}^{2} + \frac{1}{ {x}^{2} } + 2$


$36 = {x}^{2} + \frac{1}{ {x}^{2} } + 2$


$36 - 2 = {x}^{2} + \frac{1}{ {x}^{2} }$


$34 = {x}^{2} + \frac{1}{ {x}^{2} }$


___________________

Now ,


[ Using identity : ( a + b )³ = a³ + b³ + 3ab ( a + b ) ]



${(x + \frac{1}{x} )}^{3} = {x}^{3} + \frac{1}{ {x}^{3} } \: \: + \: \: 3x \times \frac{1}{x} (x + \frac{1}{x} )$


$( {3 - 2 \sqrt{2} \: \: + \: \:3 + 2 \sqrt{2} )}^{3} = {x}^{3} + \frac{1}{ {x}^{3} } + 3 \: (3 \: \: - \: 2 \sqrt{2} + \: 3 \: + \: 2 \sqrt{2} )$



$216 \: = \: {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times 6$


$216 = {x}^{3} + \frac{1}{ {x}^{3} } + 18$


$216 - 18 = \: {x}^{3} \: + \frac{1}{ {x}^{3} }$


$198 = {x}^{3} \: + \: \frac{1}{ {x}^{3} }$


Hence ,

${x}^{2} + \: \frac{1}{ {x}^{2} } = 34$


$\:198 = {x}^{3} \: + \: \frac{1}{ {x}^{3} }$