If \(x=\sqrt{3}+\sqrt{2}\), then the value of \(x^{3}-\frac{1}{x^{3}}\) is : (a) \(14 \sqrt{2}\) (b) \(14 \sqrt{3}\) (c) \(22 \sqrt{2}\) (d) \(10 \sqrt{2}\)

Given \(x = \sqrt{3} + \sqrt{2}\), to find the value of \(x^3 – \frac{1}{x^3}\), we can proceed by calculating \(x^3\) and \(\frac{1}{x^3}\) separately.

First, calculate \(x^3\):

\[
x^3 = (\sqrt{3} + \sqrt{2})^3
\]

Applying the binomial expansion, we get:

\[
(\sqrt{3})^3 + 3(\sqrt{3})^2(\sqrt{2}) + 3(\sqrt{3})(\sqrt{2})^2 + (\sqrt{2})^3
\]

\[
= 3\sqrt{3} + 3 \times 3 \sqrt{2} + 3 \times 2\sqrt{3} + 2\sqrt{2}
\]

\[
= 3\sqrt{3} + 9\sqrt{2} + 6\sqrt{3} + 2\sqrt{2}
\]

\[
= 9\sqrt{3} + 11\sqrt{2}
\]

Now, for \(\frac{1}{x^3}\), consider \(\frac{1}{x}\) first. Knowing that \(x = \sqrt{3} + \sqrt{2}\), we find the conjugate to rationalize the denominator for \(\frac{1}{x}\):

\[
\frac{1}{\sqrt{3} + \sqrt{2}} = \frac{\sqrt{3} – \sqrt{2}}{(\sqrt{3} + \sqrt{2})(\sqrt{3} – \sqrt{2})}
\]

\[
= \frac{\sqrt{3} – \sqrt{2}}{3 – 2}
\]

\[
= \sqrt{3} – \sqrt{2}
\]

Thus, \(\frac{1}{x} = \sqrt{3} – \sqrt{2}\), and hence \(\frac{1}{x^3} = (\sqrt{3} – \sqrt{2})^3\).

To simplify this, apply the binomial expansion similarly:

\[
(\sqrt{3} – \sqrt{2})^3 = (\sqrt{3})^3 – 3(\sqrt{3})^2(\sqrt{2}) + 3(\sqrt{3})(\sqrt{2})^2 – (\sqrt{2})^3
\]

\[
= 3\sqrt{3} – 3 \times 3 \sqrt{2} + 3 \times 2\sqrt{3} – 2\sqrt{2}
\]

\[
= 9\sqrt{3} – 11\sqrt{2}
\]

Now, to find \(x^3 – \frac{1}{x^3}\):

\[
x^3 – \frac{1}{x^3} = (9\sqrt{3} + 11\sqrt{2}) – (9\sqrt{3} – 11\sqrt{2})
\]

\[
= 9\sqrt{3} + 11\sqrt{2} – 9\sqrt{3} + 11\sqrt{2}
\]

\[
= 22\sqrt{2}
\]

Therefore, the value of \(x^3 – \frac{1}{x^3}\) is \(22\sqrt{2}\), making the correct answer:

(c) \(22 \sqrt{2}\).