\( \Large x^{3}-\frac{1}{x^{3}}=14 \)
=> \( \Large \left(x-\frac{1}{x}\right)^{3}+3x \times \frac{1}{x} \left(x-\frac{1}{x}\right)=14 \)
Put \( \Large x-\frac{1}{x}=z \)
Therefore, \( \Large z^{3}+3z-14=0 \)
Now, 2 = 2 satisfies the equation, hence (z - 2) factor. i.e.
z = 2
Therefore, \( \Large x-\frac{1}{x} = 2 \)