1. Solve the equation $\left(2^{x}-4\right)^{3}+\left(4^{x}-2\right)^{3}=\left(4^{x}+2^{x}-6\right)^{3}$.
III/1. Let's write $a=2^{x}$ and transform the given equation. Using the formula for the factorization of cubes, we have $(a-4)^{3}+\left(a^{2}-2\right)^{3}=\left(a^{2}+a-6\right)\left(a^{4}-a^{3}+a^{2}-6 a+12\right)$. Then, $0=(a-4)^{3}+\left(a^{2}-2\right)^{3}-\left(a^{2}+a-6\right)^{3}=\left(a^{2}+a-6\right)\left(-3 a^{3}+12 a^{2}+6 a-24\right)=-3(a+3)(a-2)(a-4)\left(a^{2}-2\right)=$ $-3(a+3)(a-2)(a-4)(a-\sqrt{2})(a+\sqrt{2})$. The only positive solutions are $a=2, a=4$ and $a=\sqrt{2}$, which give $x=1, x=2$ and $x=\frac{1}{2}$, respectively.

Substitution $a=2^{x}$: 1 point. Transforming the equation to $a^{5}-3 a^{4}-12 a^{3}+30 a^{2}+20 a-48=0$: 1 point. Factorization in the form $(a+3)(a-2)(a-4)(a-\sqrt{2})(a+\sqrt{2})$: 2 points. Positive solutions $a=2, a=4$ and $a=\sqrt{2}$: 1 point. All solutions $x=1, x=2$ and $x=\frac{1}{2}$: 2 points. If the contestant guesses one or two solutions: 1 point. If the contestant guesses all three solutions without proving there are no others: 2 points.