If $x=3+\sqrt{8}$, then the value of $x^3+\frac{1}{x^3}$ is

Given $x=3+\sqrt{8}$. Find its reciprocal $\frac{1}{x}$ by rationalizing the denominator:
$$\frac{1}{x}=\frac{1}{3+\sqrt{8}}\times \frac{3-\sqrt{8}}{3-\sqrt{8}}=\frac{3-\sqrt{8}}{3^2-(\sqrt{8}{)}^2}=\frac{3-\sqrt{8}}{9-8}=3-\sqrt{8}$$
Add $x$ and $\frac{1}{x}$:
$$x+\frac{1}{x}=(3+\sqrt{8})+(3-\sqrt{8})=6$$
Using the algebraic identity for $x^3+\frac{1}{x^3}$:
$$x^3+\frac{1}{x^3}={\left(x+\frac{1}{x}\right)}^3-3\left(x+\frac{1}{x}\right)$$
$$x^3+\frac{1}{x^3}=(6{)}^3-3(6)=216-18=198$$