Question

Please help me and do it in paper

If x=3+2√2 find the value of (√x-1/√x)^3

Answer 1

Answer:

${\large{\overbrace{\underbrace{\purple{your \: answer: - 128 \sqrt{2} }}}}} \\ \\ \red♡ dear *----* \\ </p><p></p><p>Step-by-step \: explanation: \\ </p><p> \bf\ x = 3 - 2\sqrt{2} \\ \\ \bf\ </p><p> \implies : \frac{1}{x} = \frac{1}{3 - 2 \sqrt{2} } \\ \\ \bf\ \implies: \frac{1}{x} = \frac{1(3 + 2 \sqrt{2} )}{(3 - 2 \sqrt{2} )(3 + 2 \sqrt{2} )} \\ \\ \bf\ \implies: \frac{1}{x} = \frac{3 + 2 \sqrt{2} }{ {3}^{2} - ({2 \sqrt{2} })^{2} } \\ \\ \bf\ \implies: \frac{1}{x} = \frac{3 + 2 \sqrt{2} }{9 - 8} \\ \\ \bf\ \implies: \frac{1}{x} = 3 + 2 \sqrt{2} \\ \\ \bf\ \implies: {(x - \frac{1}{x}) }^{3} ={((3 - 2 \sqrt{2}) -( 3 + 2 \sqrt{2} ))}^{3} \\ \\\bf\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies: - {(4 \sqrt{2}) }^{3} = - 128\sqrt{2} \\ \\ \large\boxed{ \fcolorbox{red}{lime}{hope \: it \: help \: you}}$

Answer 2

Answer:

8

Step-by-step explanation:

${\underline{{\text{Given}}}}$

$x = 3 + 2 \sqrt{2}$

${\underline{{\text{To Find:}}}}$

$\sqrt{x} - \dfrac{1}{ \sqrt{x} }$

${\underline{{\text{Solution:}}}}$

First we should find the value of $\sqrt{x} \text{and} \frac{1}{\sqrt{x}}$

So,

$x = 3 + 2 \sqrt{2} \\ \\ = 2 + 2 \sqrt{2} + 1 \\ \\ = ( \sqrt{2} ) 7^{2} + 2. \sqrt{2} .1 + {(1)}^{2} \\ \\ = ( \sqrt{2} + 1)^{2} \\ \\ \therefore \sqrt{x} = \sqrt{2} + 1$

Again,

$\sqrt{x} = \sqrt{2} + 1 \\ \\ \frac{1}{ \sqrt{x} } = \frac{1}{ \sqrt{2} + 1} \\ \\ [\text{Rationalizing the denominator}] \\ \\ = \frac{( \sqrt{2} - 1) }{( \sqrt{2} + 1)( \sqrt{2} - 1) } \\ \\ \frac{ \sqrt{2} - 1 }{ ({ \sqrt{2} )}^{2} - {(1)}^{2} } \\ \\ \frac{ \sqrt{2} - 1}{2 - 1} \\ \\ \frac{ \sqrt{2} - 1 }{1} \\ \\ \therefore \frac{1}{ \sqrt{x} } = \sqrt{2} - 1$

Now,

$( \sqrt{x} - \frac{1}{ \sqrt{x} } ) {}^{3} \\ \\ \{ ( \sqrt{2} + 1 ) - ( \sqrt{2} - 1)\} {}^{3} \\ \\ {( \sqrt{2} + 1 - \sqrt{2} + 1) }^{3} \\ \\ {(2)}^{3} \\ \\ 8$

Hence, The required answer is 8.

${\underline{{\text{Algebra Formula}}}}$

${(x + y)}^{2}={x}^{2}+{y}^{2}+2xy\\ \\{(x - y)}^{2}={x}^{2}+{y}^{2}-2xy\\ \\{(x+y)}^{2}= (x - y) {}^{2}+4xy\\ \\{(x-y)}^{2}=(x+y){}^{2}-4xy\\ \\ (x + y)^{2}+(x-y)^{2}=2( {x}^{2}+{y}^{2} )\\ \\(x+y)^{2}- (x-y) {}^{2}=4xy\\ \\ {(x + y)}^{3}={x}^{3}+{y}^{3}+ 3xy(x + y) \\ \\(x - y)^{3}={x}^{3}-{y}^{3}- 3xy(x - y)$