Question:
If \[ x+\frac{1}{x}=3, \] then \[ x^6+\frac{1}{x^6}= \]
(a) 927
(b) 414
(c) 364
(d) 322
Solution:
First find \[ x^2+\frac{1}{x^2} \]
Using identity:
\[ \left(x+\frac{1}{x}\right)^2 = x^2+\frac{1}{x^2}+2 \]
\[ (3)^2 = x^2+\frac{1}{x^2}+2 \]
\[ 9 = x^2+\frac{1}{x^2}+2 \]
\[ x^2+\frac{1}{x^2} = 7 \]
Now find \[ x^3+\frac{1}{x^3} \]
Using identity:
\[ \left(x+\frac{1}{x}\right)^3 = x^3+\frac{1}{x^3} + 3\left(x+\frac{1}{x}\right) \]
\[ (3)^3 = x^3+\frac{1}{x^3}+3(3) \]
\[ 27 = x^3+\frac{1}{x^3}+9 \]
\[ x^3+\frac{1}{x^3} = 18 \]
Now using identity:
\[ \left(x^3+\frac{1}{x^3}\right)^2 = x^6+\frac{1}{x^6}+2 \]
\[ (18)^2 = x^6+\frac{1}{x^6}+2 \]
\[ 324 = x^6+\frac{1}{x^6}+2 \]
\[ x^6+\frac{1}{x^6} = 324-2 \]
\[ =322 \]
Hence, the correct answer is:
\[ \boxed{322} \]