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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.

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That would be too good to be true: $a_6 - a_5 = 91 = 7 \times 13$.

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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.

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$(n+1)^3 - n^3 = 3 n^2 + 3 n + 1$. 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.

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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:

$(n^3 + 3n^2 + 3n +1 - n^3) - [n^3 - (n^3 - 3n^2 +3n -1)]\\ 6n$

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  • $\begingroup$ The algebra helped $\endgroup$ – K. Gibson Aug 30 '16 at 20:07
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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}

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