@ Haran. OP@Haran enquired about case for $n=6$.
Well lately Seiji Tomita has given a numerical solution for it.
He used the equation below to arrive at a numerical solution.
$(m^2+m+1)^3+(m^2-m+1)^3+(2)^3=(m^2+5)(m^2+m+1)(2)(m^2-m+1)$: $$ (m^2+m+1)^3+(m^2-m+1)^3+(2)^3=(m^2+5)(m^2+m+1)(2)(m^2-m+1). $$
To arrive at, $n=6$ we substitute $(m=1)$$m=1$. The downside of this is that,
when using the above equation one can get only one numerical solution
for $n=6$. The equation gives us:
$3^3+1^3+2^3=6(3*2*1)$
Tomita $$ 3^3+1^3+2^3=6(3\cdot2\cdot1). $$ Tomita has given a general method through which solutions for
different 'n'$n$ can be arrived at. His "Computational number theory"
web page is listed below:
http://www.maroon.dti.ne.jp/fermat/eindex.html
And, click On article # 456"Computational number theory" web page discusses this in #456. $x^3 + y^3 + z^3 = n x y z$.
