### 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 it involves some algebra and exponent rules.
I notice that the left-hand side of the equation has two terms that are cubes of binomials, and the right-hand side is also a cube of a binomial.
This makes me think of the identity $(a+b)^3=a^3+b^3+3ab(a+b),$ which can be used to expand or factor the sum or difference of two cubes.
Maybe I can use this identity to simplify the equation and find the solutions.
Let me try to apply the identity to the left-hand side first.
I get $(2^x-4)^3+(4^x-2)^3=(2^x-4+4^x-2)^3-3(2^x-4)(4^x-2)(2^x-4+4^x-2).$
This looks a bit messy, but maybe I can simplify it further.
I notice that the first term on the right-hand side is the same as the original right-hand side of the equation, so I can cancel them out.
This leaves me with $0=-3(2^x-4)(4^x-2)(4^x+2^x-6).$
This looks much better, because now I can use the zero product property to find the solutions.
I just need to find the values of $x$ that make any of the factors equal to zero.
I can start with the first factor, $2^x-4.$
When is this equal to zero?
I can rewrite it as $2^x=4,$ and then use logarithms to solve for $x.$
I get $x=\log_2(4),$ which is the same as $x=2,$ since $2^2=4.$
So $x=2$ is one solution.
What about the second factor, $4^x-2?$
When is this equal to zero?
I can rewrite it as $4^x=2,$ and then use logarithms again to solve for $x.$
I get $x=\log_4(2),$ which is the same as $x=\frac{1}{2},$ since $4^{1/2}=2.$
So $x=\frac{1}{2}$ is another solution.
What about the third factor, $4^x+2^x-6?$
When is this equal to zero?
I can try to rewrite it as $4^x+2^x=6,$ and then use some exponent rules to solve for $x.$
I notice that $4^x=(2^2)^x=2^{2x},$ so I can substitute that in and get $2^{2x}+2^x=6.$
This looks like a quadratic equation in terms of $2^x,$ so I can try to factor it.
I get $(2^x+3)(2^x-2)=0.$
Now I can use the zero product property again and find the values of $x$ that make either factor equal to zero.
For the first factor, $2^x+3=0,$ I can rewrite it as $2^x=-3,$ and then use logarithms to solve for $x.$
I get $x=\log_2(-3),$ but this is not a real number, since there is no power of $2$ that gives a negative number.
So this factor does not give me any real solutions.
For the second factor, $2^x-2=0,$ I can rewrite it as $2^x=2,$ and then use logarithms to solve for $x.$
I get $x=\log_2(2),$ which is the same as $x=1,$ since $2^1=2.$
So $x=1$ is another solution.
Therefore, the three real solutions are $x=2,$ $x=\frac{1}{2},$ and $x=1.$