Question 939169
<pre>
Here's the way I did it.  Not with a difference table.

Let the nth term be a<sub>n</sub>.

I noticed first that the last digits go 8,3,8,3,8,3,8.

Since the last digits differ by ±5, I noticed that if 
we subtract 3 from each one, the last digits of the new 
sequence will go 5,0,5,0,5,0 and therefore all be 
multiples of 5.

Let's subtract 3 from each, and we get a sequence
which has nth term {{{a[n]-3}}} 

5, 40, 135, 320, 625, 1080, 1715

Let's divide them by 5 and get a sequence with 
nth term {{{(a[n]-3)/5}}}

1, 8, 27, 64, 125, 216, 343

Aha. that is the sequence of cubes which has
nth term {{{n^3}}}.

So we have the equation 

{{{(a[n]-3)/5}}}{{{""=""}}}{{{n^3}}}

{{{a[n]-3}}}{{{""=""}}}{{{5n^3}}}

{{{a[n]}}}{{{""=""}}}{{{5n^3+3}}}

Edwin</pre>