Here you need to start with
a(n)
=
1 + 2 + 3 + ... + n
=
 1
2
 n(n+1)
b(n)
=
12 + 22 + 32 + ... + n2
=
 1
6
 n(n+1)(2n + 1)
c(n)
=
13 + 23 + 33 + ...? + n3
=
 1
4
 n2(n+1)2.

These are well known results found in many text books and you will find proofs in some recent problems on this website. See Picture Story and Natural Sum. You can now crack this Tough Nut because it only requires elementary algebra to eliminate n between these expressions (taken in pairs) to show that c=a2, 9b2 = 8a3 + a2 and 16c2 = 9b2 − c.