Question

If $x+\frac{1}{x}=3$, find the value $x^6+\frac{1}{x^6}$.

Answer 1

$x+\frac{1}{x}=3$,

Squaring both sides,

${(x+\frac{1}{x})}^2=(3{)}^2\Rightarrow x^2+\frac{1}{x^2}+2=9\Rightarrow x^2+\frac{1}{x^2}=9-2=7\text{Cubing both sides},{(x^2+\frac{1}{x^2})}^3=(7{)}^3\Rightarrow x^6+\frac{1}{x^6}+3(x^2+\frac{1}{x^2})=343\Rightarrow x^6+\frac{1}{x^6}+3\times 7=343\Rightarrow x^6+\frac{1}{x^6}+21=343\Rightarrow x^6+\frac{1}{x^6}=343-21=322\therefore x^6+\frac{1}{x^6}=322$