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.

**Answer**

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 \quad2^3\quad 3^3\quad 4^3\quad \ldots

$$

Note that $x^3-(x-1)^3 = 3x^2-3x+1$. So we’ll get a new sequence at the bottom:

$$

7 \quad 19\quad 37\quad 61\quad \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 \quad6\quad 12\quad 18\quad\ldots

$$

Now, note that $6(x-1)-6((x-1)-1) = 6$!

Now, the new sequence is:

$$

6\quad 6\quad 6\quad 6\quad 6\quad …

$$

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.

**Attribution***Source : Link , Question Author : blue-sky , Answer Author : Sarvesh Ravichandran Iyer*