Is this a new method for finding powers?

To test your hypothesis you could work out the form of the differences from the first few cases. begin{align*} 1^{3}-0^{3}&=1\ 2^{3}-1^{3}&=7\ 3^{3}-2^{3}&=19\ 4^{3}-3^{3}&=37 end{align*} For example rewrite out $(37-19)-(19-7)=18-6=6$ as: begin{align*} {(4^{3}-3^{3})-(3^{3}-2^{3})}&-{(3^{3}-2^{3})-(2^{3}-1^{3})}\ &=(4^{3}-2cdot3^{3}+2^3)-(3^{3}-2cdot2^{3}+1^{3})\ &=4^{3}-3cdot3^{3}+3cdot2^3-1^{3}qquad (star)\ &=6 end{align*} So you have to find the difference of two differences to get to $6$ (this is called a finite difference pattern, and you have to iterate twice to get the result of $6$ for all such differences, any further iteration ending in a $0$). Now check that pattern $(star)$ holds in general for some integer $kge3$: begin{align*} k^{3}&-3cdot(k-1)^{3}+3cdot(k-2)^3-(k-3)^{3}\ &=k^{3}-3(k^2-3k^2+3k-1) +3(k^3-2cdot3k^2+2^2cdot3k-2^3) -(k^3-3cdot3k^2+3^2cdot3k-3^3)\ &= k^3\ & -3k^3 + 9k^2 - 9k + 3\ & +3k^3-18k^2+36k-24\ & -k^3 + 9k^2-27k+27\ &=6 end{align*}

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: $ newcommandlfrac[2]{{largefrac{#1}{#2}}} $

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

$ = 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!

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 $Delta^3$.

EDIT: Let me add that there is yet another identity here which resembles Taylor series, namely, $$ f(n-1) = f(n) - Delta f(n) + Delta^2 f(n) - Delta^3 f(n). $$

Altogether, at least when $f$ is a polynomial, we thus have the following Taylor series-like formulas: begin{align*} f(n-1) &= sum_{i = 0}^infty (-1)^i [Delta^i f](n) \ f(n+1) &= sum_{i = 0}^infty [Delta^i f](n - i) \ f(n+x) &= sum_{i=0}^infty binom{x}{i} [Delta^i f](n). end{align*}

It's not something new, but for your discovery I applaud. This procedure is called the method of successive differences, and you can show that for every power the successive difference appears.

Let us say you have a sequence: $$ 1^3 quad2^3quad 3^3quad 4^3quad ldots $$

Note that $x^3-(x-1)^3 = 3x^2-3x+1$. So we'll get a new sequence at the bottom: $$ 7 quad 19quad 37quad 61quad ldots $$ Now, note that $3x^2-3x+1-(3(x-1)^2-3(x-1)+1) = 6(x-1)$. Hence, we'll get another series: $$ 0 quad6quad 12quad 18quadldots $$ Now, note that $6(x-1)-6((x-1)-1) = 6$! Now, the new sequence is: $$ 6quad 6quad 6quad 6quad 6quad ... $$ So $6$ appears as the final difference! This shows the power of algebra. As an exercise, do this for $x^4$. See the pattern of the number at the end, and if you can say something for $x^n$.

The reason, as you can see, is that at each line above, the degree of the polynomial $f(x)-f(x-1)$ decreases by $1$. Hence, at the end of three lines, you are only going to get a constant polynomial.