### Instruction: Find all real $x$ such that $(2^x-4)^3+(4^x-2)^3=(4^x+2^x-6)^3.$

Enter all the solutions, separated by commas.### Response: I notice that the equation has the form $a^3+b^3=c^3$, where $a=2^x-4$, $b=4^x-2$, and $c=4^x+2^x-6$.
I also notice that $a+b=c$, since $2^x-4+4^x-2=4^x+2^x-6$.
This means that the equation is equivalent to $a^3+b^3=(a+b)^3$, which can be simplified by expanding the right-hand side and canceling out terms.
I get $a^3+b^3=a^3+3a^2b+3ab^2+b^3$, which simplifies to $3a^2b+3ab^2=0$.
I can factor out $3ab$ from this equation, and get $3ab(a+b)=0$.
Since $a+b=c$, this is equivalent to $3abc=0$.
This means that either $a=0$, $b=0$, or $c=0$.
I can solve for $x$ in each case by setting $a=0$, $b=0$, or $c=0$ and using logarithms.
If $a=0$, then $2^x-4=0$, so $2^x=4$, and $x=2$.
If $b=0$, then $4^x-2=0$, so $4^x=2$, and $x=1/2$.
If $c=0$, then $4^x+2^x-6=0$, which is a quadratic equation in $2^x$.
I can let $u=2^x$, and rewrite the equation as $u^2+u-6=0$.
I can factor this equation as $(u+3)(u-2)=0$, and get $u=-3$ or $u=2$.
Since $u=2^x$, I can only use the positive solution, and get $2^x=2$, and $x=1$.
So the solutions are $x=2$, $x=1/2$, and $x=1$.