$u^3+v^3+3^{5}w^3=2\cdot3^{2}uvw$ has no non- trivial solutions if $u,v,w$ are pairwise co-prime.

Question

Let $u,v,w$ be 3 pairwise coprime integers. Then $$u^3+v^3+3^{5}w^3=2\cdot3^{2}uvw$$ has no non-trivial solutions. How can I prove this?

I have tried to consider many individual cases such as $uvw>0$,$uvw<0$, $max(u,v,w)=u$ etc. a pretty tedious approach. I am certain there must be simpler ones. Any hints?

Answer 1

$$ u = 1, \; \; v = -1, \; \; w = 0 $$

Answer 2

$u = -v, w = 0$.

Generalizing Will Jagy's.

Don't think that either of these helps much.