Question

Maths experts please help..
If x = 3+2√2 find the value of x^3 + 1/x^3.

Answer 1

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Answer 2

$\large\underline\mathfrak{Answer-}$

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

$\large\underline\mathfrak{Explanation-}$

Given : x = 3+2√2

To find : Value of ${x}^{3} + \frac{1}{ {x}^{3} }$

Solution : x = 3+√2

$\dfrac{1}{x}$ = 1/(3+2√2)

Rationalize it :

=> 1/(3+2√2) × (3-2√2)/(3-2√2)

=> (3-2√2)/(3²-2√2²) [ a² - b² = (a+b)(a-b) ]

=> 3-2√2

Now,

We know that,

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

Put the values,

=> ( 3 + 2√2 + 3 - 2√2 ) [ ( 3+2√2 )² + ( 3-2√2 )² - 1 ]

=> 6 ( 9 + 8 + 12√2 + 9 + 8 - 12√2 - 1 )

=> 6 ( 33 )

=> 198

$\therefore$ ${x}^{3}+\dfrac{1}{ {x}^{3} }=198$