If $x = \sqrt{3} + \sqrt{2},$ $x = \sqrt 3 {\text{ + }}\sqrt 2 {\text{,}}$ then the value of $x^{3} - \frac{1}{x^{3}}$ ${x^3} - \frac{1}{{{x^3}}}$ is?

1 $10 \sqrt{2}$ $10\sqrt 2$
2 $14 \sqrt{2}$ $14\sqrt 2$
3 $22 \sqrt{2}$ $22\sqrt 2$
4 $8 \sqrt{2}$ $8\sqrt 2$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} x = \sqrt{3} + \sqrt{2} \\ \frac{1}{x} = \frac{1}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} \\ \frac{1}{x} = \sqrt{3} - \sqrt{2} \\ x^{3} - \frac{1}{x^{3}} \\ = {[x - \frac{1}{x}]}^{3} + 3 \times x \times \frac{1}{x} (x - \frac{1}{x}) \\ = {(\sqrt{3} + \sqrt{2} - \sqrt{3} + \sqrt{2})}^{3} + 3 (\sqrt{3} + \sqrt{2} - \sqrt{3} + \sqrt{2}) \\ = {(2 \sqrt{2})}^{3} + 3 (2 \sqrt{2}) \\ = 16 \sqrt{2} + 6 \sqrt{2} \\ = 22 \sqrt{2} \end{aligned}

$\eqalign{ & x = \sqrt 3 + \sqrt 2 \cr & \frac{1}{x} = \frac{1}{{\sqrt 3 + \sqrt 2 }} \times \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & \frac{1}{x} = \sqrt 3 - \sqrt 2 \cr & {x^3} - \frac{1}{{{x^3}}} \cr & = {\left[ {x - \frac{1}{x}} \right]^3} + 3 \times x \times \frac{1}{x}\left( {x - \frac{1}{x}} \right) \cr & = {\left( {\sqrt 3 + \sqrt 2 - \sqrt 3 + \sqrt 2 } \right)^3} + 3\left( {\sqrt 3 + \sqrt 2 - \sqrt 3 + \sqrt 2 } \right) \cr & = {\left( {2\sqrt 2 } \right)^3} + 3\left( {2\sqrt 2 } \right) \cr & = 16\sqrt 2 + 6\sqrt 2 \cr & = 22\sqrt 2 \cr}$