Do the differences of perfect squares apply to perfect cubes and more?

I'm curious about a special property of squares. The difference between perfect squares starting from 0 are 1,3,5,7,9..., where each difference goes up by 2. I want to know if there are any patterns for perfect cubes or quartics. Are the differences in a pattern of some sort?

Solution 1:

For cubes $$\begin {array} &0&&1&&8&&27&&64&&125\\&1&&7&&19&&37&&61\\&&6&&12&&18&&24\\&&&6&&6&&6 \end {array}$$ Try it yourself for quartics. What do you find?

Solution 2:

Look at the pattern of the differences of the differences and so on :)

Solution 3:

In general, the sequence is generated by a polynomial of degree $d$, then the sequence of the first differences is generated by a polynomial of degree $d-1$.

To see this, consider a polynomial $p(n)$ with leading coefficient $a$. It is easy to see that $p(n+1)$ also has degree $d$ and leading coefficient $a$. If the term of degree $d-1$ in $p(n)$ is $b$, then the corresponding coefficient in $p(n+1)$ is $b+ad$. Hence $q(n)=p(n+1)-p(n)$ has degree $d-1$ with leading coefficient $ad$.

As an example with the cubes ($n\geq 0$):

$p_0(n)=n^3$: generates $0,1,8,27,64,125,\ldots$

$p_1(n)=p_0(n+1)-p_0(n)=3n^2+3n+1$: generates $1,7,19,37,61,\ldots$

$p_2(n)=p_1(n+1)-p_1(n)=6n+6$: generates $6,12,18,24,\ldots$

$p_3(n)=p_2(n+1)-p_2(n)=6$: generates $6,6,6,\ldots$