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If $ x-\frac{1}{x}=3+2 \sqrt{2} $, find the value of $ x^{3}- \frac{1}{x^{3}} $.
Given:
\( x-\frac{1}{x}=3+2 \sqrt{2} \)
To do:
We have to find the value of \( x^{3}- \frac{1}{x^{3}} \).
Solution:
$x-\frac{1}{x}=3+2 \sqrt{2}$
Cubing both sides, we get,
$(x-\frac{1}{x})^{3}=(3+2 \sqrt{2})^{3}$
$x^{3}-\frac{1}{x^{3}}-3\times x \times \frac{1}{x}(x-\frac{1}{x})=(3)^{3}+(2 \sqrt{2})^{3}+3 \times 3 \times 2 \sqrt{2}(3+2 \sqrt{2})$
$x^{3}-\frac{1}{x^{3}}-3 \times(3+2 \sqrt{2})=27+16 \sqrt{2}+18 \sqrt{2}(3+2 \sqrt{2})$
$x^{3}-\frac{1}{x^{3}}-3 \times(3+2 \sqrt{2})=27+16 \sqrt{2}+54 \sqrt{2}+72$
$x^{3}+\frac{1}{x^{3}}-3(3+2 \sqrt{2})=99+70 \sqrt{2}$
$x^{3}+\frac{1}{x^{3}}=99+70 \sqrt{2}+9+6 \sqrt{2}$
$x^{3}+\frac{1}{x^{3}}=108+76 \sqrt{2}$
The value of $x^{3}+\frac{1}{x^{3}}$ is $108+76 \sqrt{2}$.
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