Find all values x such that $\displaystyle (2^x-4)^3+(4^x-2)^3 = (4^x+2^x-6)^3$

I expanded the entire thing:

$\displaystyle 2^3^x+2^2^x-38+4^3^x+4^2^x+4^x = (4^3^x)(2^5^x)+(4^2^x)(2^3^x)+4^x+4^x-105+2^x+2^x$

but not quite sure what to do from here. Is there a logarithm I can use? Or would I have to go further and get it all to one side equaling to 0?