If x = 3 + 2 , then the value of x^3 - 1x^3 is?

Answer: Option C

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