Question

If $x+\frac{1}{x}=7$, find $x^3+\frac{1}{x^3}$.

Answer 1

It is given that $x+\frac{1}{x}=7$, by taking cubes on both sides we get:

${(x+\frac{1}{x})}^3=7^3\Rightarrow {\left(x\right)}^3+{\left(\frac{1}{x}\right)}^3+(3\times x\times \frac{1}{x})(x+\frac{1}{x})=343(\because (a+b{)}^3=a^3+b^3+3ab(a+b))\Rightarrow x^3+\frac{1}{x^3}+3\left(7\right)=343(Givenx+\frac{1}{x}=7)\Rightarrow x^3+\frac{1}{x^3}+(3\times 7)=343$

$\Rightarrow x^3+\frac{1}{x^3}+21=343\Rightarrow x^3+\frac{1}{x^3}=343-21\Rightarrow x^3+\frac{1}{x^3}=322$

Hence, $x^3+\frac{1}{x^3}=322$.