Question

ANYONE PLS ANSWER IT FASTLY.......​

Answer 1

Answer:

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

Now,

$x - \frac{1}{x} = 3 + 2 \sqrt{2} - 3 + 2 \sqrt{2} \\ \\ x - \frac{1}{x} = 4 \sqrt{2} \\ \\ \sf cubing \: both \: sides - \\ \\ (x - \frac{1}{x} )^{3} = (4 \sqrt{2} ) {}^{3} \\ \\ x {}^{3} - \frac{1}{x {}^{3} } - 3(x - \frac{1}{x} ) = 128 \sqrt{2} \\ \\ x {}^{3} - \frac{1}{x {}^{3} } - 3(4 \sqrt{2} ) = 128 \sqrt{2} \\ \\ x {}^{3} - \frac{1}{x {}^{3} } = 128 \sqrt{2} + 12 \sqrt{2} \\ \\ \boxed{ \bf x {}^{3} - \frac{1}{x {}^{3} } = 140 \sqrt{2} }$

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Answer 2

Heyaa☺

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Your answer is in the above attachment.

Refer to it.

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Hope it helps...✌