Question

if x=3+2 root2 find [x-1/x]whole cube

Answer 1

Given -

x = 3 + 2√2.

To find -

Find the value of ( x - 1 / x )³.

Solution -

x = 3 + 2√2 .

( 1 / x )

=> [ 1 / 3 + 2√2 ]

=> [ 1 / 3 + 2√2 ] × [ 3 - 2√2 / 3 - 2√2 ]

=> [ 3 - 2√2 ] / [ ( 3 + 2√2 ) × ( 3 - 2√2 ) ]

=> [ 3 - 2√2 ] / [ 9 - 8 ]

=> 3 - 2√2 .

Thus ,

( 1 / x ) = 3 - 2√2

[ x - 1 / x ]

=> [ 3 + 2√2 ] - [ 3 - 2√2 ]

=> 3 + 2√2 - 3 + 2√2

=> 4√2 .

[ x - 1 / x ]³

=> { 4 √ 2 }³

=> 64 × 2√2

=> 128√2 .

This is the required answer .

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Additional Information -

( a + b )² = a² + 2ab + b²

( a - b )² = a² - 2ab + b²

( a + b )( a - b ) = a² - b²

( a + b )³ = a³ + 3ab ( a + b ) + b³

( a - b )³ = a³ - 3ab ( a + b ) - b³

( a + b + c )³ = a³ + b³ + c³ + 3 ( a + b )( b + c )( c + a )

a³ + b³ + c³ - 3abc = ( a + b + c )( a² + b² + c² - ab - bc - ca )

When a + b + c = 0 ,

a³ + b³ + c³ = 3abc .

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Answer 2

ANSWER✔

$\large\underline\bold{GIVEN,}$

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

$\sf\dashrightarrow \bigg[ x-\dfrac{1}{x} \bigg]^3$

$\large\underline\bold{TO\:FIND,}$

$\sf\dashrightarrow the\:the\:value\:of\:\bigg[x- \dfrac{1}{x} \bigg]^3$

FORMULA IN USE,

$\large{\boxed{\bf{ (x+y)^2=x^2+y^2+2xy}}}$

$\large\underline\bold{SOLUTION,}$

SOLVING THE GIVEN EQUATION,

$\sf\therefore \bigg[ x \: -\: \dfrac{1}{x}$

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

$\sf\large\therefore SUBSTITUTING\:THE\:VALUE\:OF\:'X'\:IN \:THE\:GIVEN \:EQUATION$

$\sf\implies (3+2 \sqrt{2}) - \dfrac{1}{ 3+2 \sqrt{2}}$

$\sf\implies \dfrac{(3+2 \sqrt{2})}{1} - \dfrac{1}{ 3+2 \sqrt{2}}$

$\sf\implies \dfrac{(3+2 \sqrt{2})^2-1}{(3+2 \sqrt{2})}$

$\sf\implies \dfrac{(3)^2+(2 \sqrt{2})^2+2(3)(2 \sqrt{2})-1}{3+2\sqrt{2}}$

$\sf\implies \dfrac{9+(8)+12 \sqrt{2}-1}{3+2\sqrt{2}}$

$\sf\implies \dfrac{17-1 +12 \sqrt{2}}{3+2\sqrt{2}}$

$\sf\implies \dfrac{16 +12 \sqrt{2}}{3+2\sqrt{2}}$

$\sf\implies \dfrac{4 \sqrt{2} (3+2\sqrt{2})}{3+2\sqrt{2}}$

$\sf\implies \dfrac{4 \sqrt{2} \cancel{ (3+2\sqrt{2})}}{\cancel{3+2\sqrt{2}}}$

$\sf\implies 4 \sqrt{2}$

NOW,

$\sf\implies (4 \sqrt{2})^3$

$\sf\implies 128 \sqrt{2}$

$\large{\boxed{\bf{ \star\:\: 128 \sqrt{2} \:\: \star}}}$

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