Question

If x=3+2root2 then find the value of x cube +1 by x cube

Answer 1

$x = 3 + 2 \sqrt{2} \\ \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \\ \: \: \: \: \: = \frac{3 - 2 \sqrt{2} }{1 }\\ \: \: \: \: \: = 3 - 2 \sqrt{2} \\ x + \frac{1}{x} = 6 \\ cubing \: both \: sides \\ (x + \frac{1}{x})^{3} = 216 \\ {x}^{3} + ( { \frac{1}{x}) }^{3} + 3.x. \frac{1}{x} (x + \frac{1}{x} ) = 216 \\ {x}^{3} + ( { \frac{1}{x}) }^{3} + 3(6) = 216 \\ {x}^{3} + ( { \frac{1}{x} )}^{3} = 216 - 18 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 198$
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Answer 2

Hi there!

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Given :

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

To find ;

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

Solution :

$x = 3 + 2\sqrt{2} \\ \\ \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } \\ \\ \frac{1}{x} = \frac{3 - 2 \sqrt{2} }{(3) {}^{2} - (2 \sqrt{2} ) {}^{2} } \\ \\ \frac{1}{x} = \frac{3 - 2 \sqrt{2} }{9 - 8} \\ \\ \frac{1}{x} = 3 - 2 \sqrt{2}$

And,

$x + \frac{1}{x} = 3 + 2 \sqrt{2} + 3 - 2 \sqrt{2} \\ \\ x + \frac{1}{x} = 3 + 3 \\ \\ x + \frac{1}{x} = 6$

Now,

$(x + \frac{1}{x} ) {}^{3} = (6) {}^{3} \\ \\ x {}^{3} + \frac{1}{x {}^{3} } + 3(x + \frac{1}{x} ) = 216 \\ \\ x {}^{3} + \frac{1}{x {}^{3} } + 3 \times 6 = 216 \\ \\ x {}^{3} + \frac{1}{x {}^{3} } + 18 = 216 \\ \\ x {}^{3} + \frac{1}{x {}^{3} } = 216 - 18 \\ \\ \boxed{ \red{ \bold{ x {}^{3} + \frac{1}{x {}^{3} } = 198}}}$


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Thanks for the question!

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