Description for Correct answer:
\( \Large \left(x^{2}+\frac{1}{x^{2}}\right)=\frac{17}{4} \)

=>\( \Large x^{2}+\frac{1}{x^{2}}+2-2=\frac{17}{4} \)

=>\( \Large \left(x-\frac{1}{x}\right)^{2}+2=\frac{17}{4} \)

=>\( \Large \left(x-\frac{1}{x}\right)^{2}=\frac{17}{4}-2 \)

=>\( \Large \left(x-\frac{1}{x}\right)^{2}=\frac{3}{2} \)

On cubing both sides, we get

\( \Large \left(x-\frac{1}{x}\right)^{3}= \left(\frac{3}{2}\right)^{3} \)

=>\( \Large x^{3}-\frac{1}{x^{3}}-3 \times \frac{1}{x} \times \left(x-\frac{1}{x}\right)=\frac{27}{8} \)

=>\( \Large x^{3}-\frac{1}{x^{3}}=\frac{27}{8}+3 \times \frac{3}{2} \)

=>\( \Large x^{3}-\frac{1}{x^{3}}=\frac{27}{8}+\frac{9}{2} \)

=>\( \Large \left(x^{3}-\frac{1}{x^{3}}\right)=\frac{63}{8} \)