Solution: We want to evaluate \[ \lim_{n \to \infty} \frac{1^3 + 2^3 + \cdots + n^3}{\sqrt{3n^8 + 5} }. \] Recall that \[ 1^3 + 2^3 + \cdots + n^3 = \left( \frac{n(n+1)}{2} \right) ^2. \] So we have \begin{align*} \frac{1^3 + 2^3 + \cdots + n^3}{\sqrt{3n^8 + 5} } & = \frac{n^2 (n+1)^2}{4 \sqrt{3n^8 + 5} } \\ & = \frac{n^2 \left( n^2 + 2n + 1 \right) }{4 \sqrt{n^8 \left( 3 + \frac{5}{n^8} \right) } } \\ & = \frac{n^4 \left( 1 + \frac{2n}{n^2} + \frac{1}{n^2} \right) }{4n^4 \sqrt{3+\frac{5}{n^8}} } \\ & = \frac{1+\frac{2}{n} + \frac{1}{n^2}}{4 \sqrt{3+\frac{5}{n^8}} }. \end{align*}

Thus, we got \begin{align*} \lim_{n \to \infty} \frac{1^3 + 2^3 + \cdots + n^3}{\sqrt{3n^8 + 5} } & = \lim_{n \to \infty} \frac{1+\frac{2}{n} + \frac{1}{n^2}}{4 \sqrt{3+\frac{5}{n^8}} } \\ & = \frac{1}{4\sqrt{3} }. \end{align*} Therefore, \[ \textcolor{blue}{\boxed{\lim_{n \to \infty} \frac{1^3 + 2^3 + \cdots + n^3}{\sqrt{3n^8 + 5} } = \frac{1}{4\sqrt{3} }}} \]