### 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: This problem looks like a cubic equation in disguise, so I will try to simplify it using some algebraic identities.
I notice that the expression $(4^x+2^x-6)^3$ is the sum of the other two terms, so I can use the identity $(a+b)^3=a^3+b^3+3ab(a+b)$ to expand it.
If I do that, I get $(4^x+2^x-6)^3=(2^x-4)^3+(4^x-2)^3+3(2^x-4)(4^x-2)(4^x+2^x-6).$
Now I can subtract $(2^x-4)^3+(4^x-2)^3$ from both sides and get $0=3(2^x-4)(4^x-2)(4^x+2^x-6).$
This means that either $2^x-4=0$ or $4^x-2=0$ or $4^x+2^x-6=0.$
I can solve each of these equations separately and find the possible values of $x$.
For the first equation, $2^x-4=0$, I can add $4$ to both sides and get $2^x=4.$
Then I can take the logarithm base $2$ of both sides and get $x=\log_2(4).$
I know that $\log_2(4)=2$ because $2^2=4$, so $x=2$ is one solution.
For the second equation, $4^x-2=0$, I can add $2$ to both sides and get $4^x=2.$
Then I can take the logarithm base $4$ of both sides and get $x=\log_4(2).$
I know that $\log_4(2)=\frac{1}{2}$ because $4^{\frac{1}{2}}=2$, so $x=\frac{1}{2}$ is another solution.
For the third equation, $4^x+2^x-6=0$, I can try to factor it using some trial and error.
I notice that if $x=0$, then the expression becomes $1+1-6=-4$, which is too small.
If $x=1$, then the expression becomes $4+2-6=0$, which works.
If $x=2$, then the expression becomes $16+4-6=14$, which is too big.
So $x=1$ is the only solution for this equation.
Therefore, the final answer is $x=2,\frac{1}{2},1.$