<s>[INST] Is this a new method for finding powers? Playing with a pencil and paper notebook I noticed the following: 
$x=1$
$x^3=1$

$x=2$
$x^3=8$

$x=3$
$x^3=27$

$x=4$
$x^3=64$

$64-27 = 37$
$27-8 = 19$
$8-1 = 7$

$19-7=12$
$37-19=18$

$18-12=6$

I noticed a pattern for first 1..10 (in the above example I just compute the first 3 exponents) exponent values, where the difference is always 6 for increasing exponentials. So to compute $x^3$ for $x=5$, instead of $5\times 5\times 5$, use $(18+6)+37+64 = 125$.
I doubt I've discovered something new, but is there a name for calculating exponents in this way? Is there a proof that it works for all numbers?
There is a similar less complicated pattern for computing $x^2$ values.
 [/INST] 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}-2\cdot3^{3}+2^3)-(3^{3}-2\cdot2^{3}+1^{3})\\
&=4^{3}-3\cdot3^{3}+3\cdot2^3-1^{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 $k\ge3$:
\begin{align*}
k^{3}&-3\cdot(k-1)^{3}+3\cdot(k-2)^3-(k-3)^{3}\\
&=k^{3}-3(k^2-3k^2+3k-1)
+3(k^3-2\cdot3k^2+2^2\cdot3k-2^3)
-(k^3-3\cdot3k^2+3^2\cdot3k-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*}
</s>