Question:
If
\[ x^3-\frac{1}{x^3}=14, \] then \[ x-\frac{1}{x}= \]
(a) 5
(b) 4
(c) 3
(d) 2
Solution:
Using identity:
\[ \left(x-\frac{1}{x}\right)^3 = x^3-\frac{1}{x^3} – 3\left(x-\frac{1}{x}\right) \]
Substituting the given value:
\[ \left(x-\frac{1}{x}\right)^3 = 14-3\left(x-\frac{1}{x}\right) \]
Let
\[ x-\frac{1}{x}=a \]
Then
\[ a^3=14-3a \]
\[ a^3+3a-14=0 \]
Checking the options:
\[ a=2 \]
\[ 2^3+3(2)-14 = 8+6-14 = 0 \]
Hence,
\[ x-\frac{1}{x}=2 \]
Therefore, the correct answer is:
\[ \boxed{2} \]