Question:
If \[ x+\frac{1}{x}=2, \] then \[ x^3+\frac{1}{x^3}= \]
(a) 64
(b) 14
(c) 8
(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:
\[ (2)^3 = x^3+\frac{1}{x^3}+3(2) \]
\[ 8 = x^3+\frac{1}{x^3}+6 \]
\[ x^3+\frac{1}{x^3} = 8-6 \]
\[ =2 \]
Hence, the correct answer is:
\[ \boxed{2} \]