jmc

Let $a=2^x-4$ and $b=4^x-2.$ Then $a+b=2^x+4^x-6,$ and the equation becomes $$a^3+b^3=(a+b{)}^3.$$Expanding, we get $a^3+b^3=a^3+3a^{2}b+3a{b}^2+b^3.$ Then $3a^{2}b+3a{b}^2=0,$ which factors as $$3ab(a+b)=0.$$Thus, $a=0,$ $b=0,$ or $a+b=0.$

For $a=0,$ $2^x-4=0,$ so $x=2.$

For $b=0,$ $4^x-2=0,$ so $x=\frac{1}{2}.$

For $a+b=0,$ $$2^x+4^x=6.$$Note that $x=1$ is a solution. Since $2^x+4^x$ is an increasing function, it is the only solution.

Thus, the solutions are $\boxed{\frac{1}{2},1,2}.$