Correct Answer: 322

Explanation:

$(x+\frac{1}{x})$ = 3

Now, $(x^{6}+\frac{1}{x^{6}})=(x^{2})^{3}+(\frac{1}{x^{2}})^{3}$

= $(x^{2}+\frac{1}{x^{2}})^{3}- 3\times x^{2}\times \frac{1}{x^{2}}(x^{2}+\frac{1}{x^{2}})$

= $(x^{2}+\frac{1}{x})^{3}-3(x^{2}+\frac{1}{z^{2}})$

= {$(x+\frac{1}{x})^{2}-2\times x\times \frac{1}{x}$}$^{3}$-3{${(x+\frac{1}{x})^{2}}-2\times x\times \frac{1}{x}$}

= {$(3)^{2}-2$}$^{3}$-3{(3)$^{3}$-2}

= (9-2)$^{3}$-3(9-2)

= (7)$^{3}$-3×7

=343-21=322