Find the sum of series without differentiation

Let \begin{eqnarray*} S=\sum_{i=1}^{\infty} i^2 x^i. \end{eqnarray*} Now multiply by $(1-3x+3x^2-x^3)$ and note that for $i \geq 2$ \begin{eqnarray*} (i-3)^2-3(i-2)^2+3(i-1)^2-i^3=0. \end{eqnarray*} Examine the lower order terms more carefully, and we have \begin{eqnarray*} (1-3x+3x^2-x^3)S=x(1+x) \end{eqnarray*} giving the well known formula \begin{eqnarray*} S=\sum_{i=1}^{\infty} i^2 x^i =\frac{x(1+x)}{(1-x)^3}. \end{eqnarray*}