Solutions to $8^n = a^3+b^3+c^3-3abc$ Let $n$ be a positive integer. What is the number of solutions to the equation
$$8^n = a^3+b^3+c^3-3abc$$ 
with integers $a\geq b\geq c\geq 0$?
We have the factoring
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
meaning that both factors must be powers of $2$.

Strangely enough, the solution is finite.
for the equation:
$$X^3+Y^3+Z^3-3XYZ=q=ab$$
If it is possible to decompose the coefficient as follows:
$4b=k^2+3t^2$
Then the solutions are of the form:
$$X=\frac{1}{6}(2a-3t\pm{k})$$
$$Y=\frac{1}{6}(2a+3t\pm{k})$$
$$Z=\frac{1}{3}(a\mp{k})$$
