Is this a new method for finding powers?

What you have discovered is a finite difference calculation. For any function $f$, in this case the third-power function $$ f(n) = n^3 $$ we can define the forward difference, or forward discrete derivative: $$ \Delta f(n) = f(n+1) - f(n) = 3n^2 + 3n + 1 $$ Likewise, \begin{align*} \Delta \Delta f(n) = \Delta^2 f(n) &= 6n+ 6 \\ \Delta^3 f(n) &= 6 \\ \Delta^4 f(n) &= 0. \end{align*}

Your computation, $$ 5^3 = 64 + 37 + 18 + 6 $$ is the statement $$ f(5) = f(4) + \Delta f(3) + \Delta^2 f(2) + \Delta^3 f(1), $$ or more generally $$ f(n) = f(n-1) + \Delta f(n-2) + \Delta^2 f(n-3) + \Delta^3 f(n-4). $$ This is one discrete analogue of Taylor series (the more common analogue is Newton's series). The reason it works is that, for $f(n) = n^3$, $\Delta^4$ and beyond are all zero. So the summation stops once we get to $\Delta^3$.

For a little bit more, see the answer "General method for indefinite summation" which explains how exactly this representation using forward differences allows you to easily find the formula for indefinite summation of powers. Applied to your case you get:

0,  1,  8, 27
  1,  7, 19
    6, 12
      6

and hence:

$n^3 = 0 \binom{n}{0} + 1 \binom{n}{1} + 6 \binom{n}{2} + 6 \binom{n}{3}$.

which immediately gives: $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $

$\sum_{k=0}^{n-1} k^3 = 0 \binom{n}{1} + 1 \binom{n}{2} + 6 \binom{n}{3} + 6 \binom{n}{4} = n\lfrac{n-1}{2}(1+\lfrac{n-2}{3}(6+\lfrac{n-3}{4}(6)))$

$\ = \lfrac{n^2 (n-1)^2}{4}$.

and then, if you prefer the indices to end at $n$:

$\sum_{k=1}^n k^3 = \lfrac{(n+1)^2 n^2}{4}$.

As you can see, hardly any computation was necessary to get this result!