$y=x^3$ as a sequence of primes increasing by 6n over the interval $[1,\infty)$ Lets let $a_n = (x^3 \vert x \epsilon \mathbb N)$
$a_1= 1$,
$a_2= 8$,
$a_3= 27$,
$a_4= 64$,
$a_5= 125$......
$a_2-a_1= 7$
$a_3-a_2= 19$
$a_4-a_3= 37$
$a_5-a_4= 61$
$(a_3-a_2)-(a_2-a_1)= 12$ $6n \vert n=2$
$(a_4-a_3)-(a_3-a_2)= 18$ $6n \vert n=3$
$(a_5-a_4)-(a_4-a_3)= 24$ $6n \vert n=4$ 
My guess is this continues on forever, however I have no idea to go about proving it. 
 A: That would be too good to be true: $a_6 - a_5 = 91 = 7 \times 13$.
A: You seem to be guessing that two separate things "go on forever". 
One is that the differences of consecutive cubes will always be prime. @par 's answer finds 91, so no.
The second is that the second differences increase by 6 each time. That is true, and isn't hard to prove with simple algebra.
A: $(n+1)^3 - n^3 = 3 n^2 + 3 n + 1$.  Of course this isn't always prime, e.g. it's divisible by $7$ if $n \equiv 1 \mod 7$ or $n \equiv 5 \mod 7$.  
A consequence of Bunyakovsky's conjecture is that there are infinitely many primes of this form, but we are not able to prove anything like that with current methods.
A: regarding your second observation:
$(a_{n+1} - a_{n}) - (a_n - a_{n-1})\\
((n+1)^3 -n^3) - (n^3 - (n-1)^3)$
Multiply it out, add up the pieces and simplify:
$(n^3 + 3n^2 + 3n  +1 - n^3) - [n^3 - (n^3 - 3n^2 +3n -1)]\\
6n$ 
A: In spite of your title, this doesn't have anything to do with primes.
With that said, the binomial theorem tells us that
$$(n+a)^3=n^3+3an^2+3a^2n+1.$$
Apply this twice (once with $a=1$ and once with $a=-1$) to see that
\begin{align}\left((n+1)^3-n^3\right) - \left(n^3 - (n-1)^3)\right)
&= (3n^2 + 3n + 1) - (3n^2 - 3n + 1)
\\\\&= 6n.
\end{align}
