If First of Four Numbers in Geometric Sequence is Cube then Fourth is Cube

Theorem
Let $P = \left({a, b, c, d}\right)$ be a geometric progression of integers.

Let $a$ be a cube number.

Then $d$ is also a cube number.

Proof
From Form of Geometric Progression of Integers:
 * $P = \left({k p^3, k p^2 q, k p q^2, k q^3}\right)$

for some $k, p, q \in \Z$.

If $a = k p^3$ is a cube number it follows that $k$ is a cube number: $k = r^3$, say.

So:
 * $P = \left({r^3 p^3, r^3 p^2 q, r^3 p q^2, r^3 q^3}\right)$

and so $d = r^3 q^3 = \left({r q}\right)^3$.