Merging 2 sums together

Start with $\sum_{n=0}^\infty x^n = \frac{1}{1-x}$ which converges for $|x|<1$ .

Differentiate: $\sum_{n=0}^\infty n x^{n-1} = \frac{1}{(1-x)^2}$

Multiply by $x$ : $\sum_{n=0}^\infty n x^n = \frac{x}{(1-x)^2}$

Differentiate: $\sum_{n=0}^\infty n^2 x^{n-1} = \frac{1+x}{(1-x)^3}$

Multiply by $x$ : $\sum_{n=0}^\infty n^2 x^n = \frac{x+x^2}{(1-x)^3}$

Differentiate: $\sum_{n=0}^\infty n^3 x^{n-1} = \frac{1+4x+x^2}{(1-x)^4}$

Multiply by $x$ : $\sum_{n=0}^\infty n^3 x^n = \frac{x+4x^2+x^3}{(1-x)^4}$ and since all of these steps maintained the same interval of convergence, this last series converges for $|x|<1$ as well. In particular, we can set $x=\frac{1}{3}$ on both sides of the equation to conclude

$\frac{1^3}{3^1} + \frac{2^3}{3^2} + \frac{3^3}{3^3} + \cdots = \sum_{n=0}^\infty n^3\left(\frac{1}{3}\right)^n = \frac{\left(\frac{1}{3}\right) + 4\left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^3}{\left(1-\frac{1}{3}\right)^4} = \frac{33}{8} = \boxed{4.125}$