For the purpose of this question and the patterns pointed out, $0$ is not a perfect square (some think it is, others no).
The first perfect squares are $1$, $4$, $9$, $16$, $25$ and so on. The difference between two consecutive squares, each time, increases by $2$.
$$2^2 – 1^2 = 3$$
$$3^2 – 2^2 = 5$$
$$4^2 – 3^2 = 7$$
A pattern can also be established based on the perfect square's position.
Given a perfect square's position ($1$ is the first perfect square, $4$ is the second perfect square), the next perfect square can be produced by:
$$PerfectSquare + 2(PerfectSquarePosition) + 1$$
$1$ is the first perfect square. The next perfect squares ($4$) is $1 + 2(1) + 1$. Since $4$ is the second perfect square, the perfect square after $4$ is $4 + 2(2) + 1$. The perfect square after $9$ is $16$, which equals $9 + 2(3) + 1$.
Is there such a pattern for perfect cubes? I was unable to find a way to consistently relate $1$, $8$, $27$ and so one, both through the first pattern I pointed out (the difference between two consecutive cubes) and the second pattern (finding the next perfect square based on the previous perfect square and its position).
Best Answer
$$(n+1)^3 - n^3 = 3n(n+1) +1$$
For example,
$$10^3 - 9^3 = 3\cdot 9 \cdot 10 + 1 = 271 \\ 1000 - 729 = 271 $$
That pattern looks a little like the pattern for squares...