Jim sent in this solution, using the ideas from our hints.

$a(n)=1+2+\ldots+n=\frac{n(n+1)}{2}$

$b(n)=1^2+2^2+\ldots+n^2=\frac{n(n+1)(2n+1)}{6}$

$c(n)=1^3+2^3+\ldots+n^3=\frac{n^2(n+1)^2}{4}$

It's obvious that $c=a^2$, from this.

Also, $2a=n^2+n$, so, solving the quadratic (and using the fact that $n> 0$), we get $n=\frac{-1+\sqrt{1+8a}}{2}$.

Now substitute this for $n$ in $b$, to get $b=\frac{n(n+1)(2n+1)}{6}=\frac{a}{3}\times(2n+1)=\frac{a}{3}\times\sqrt{1+ 8a}$. So $3b=a\sqrt{1+8a}$, so $9b^2=a^2+8a^3$.

Now we can combine these two expressions: $9b^2=c+8c\sqrt{c}$, so $8c\sqrt{c}= 9b^2-c$, so $64c^3=81b^4-18b^2c+c^2$.

(It's easy to check that the expressions above in terms of $n$ do work in this!)