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

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

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

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

Hence, $x^3+\frac{1}{x^3}=322$.