G1.3 Find the sum of all real roots $x$ of the equation $\left(2^{x}-4\right)^{3}+\left(4^{x}-2\right)^{3}=\left(4^{x}+2^{x}-6\right)^{3}$.
Let $a=2^{x}-4, b=4^{x}-2, a+b=4^{x}+2^{x}-6$, the equation is equivalent to $a^{3}+b^{3}=(a+b)^{3}$
$$
\begin{array}{l}
(a+b)\left(a^{2}-a b+b^{2}\right)=(a+b)^{3} \\
a^{2}-a b+b^{2}=a^{2}+2 a b+b^{2} \text { or } a+b=0 \\
3 a b=0 \text { or } 4^{x}+2^{x}-6=0 \\
2^{x}=4 \text { or } 4^{x}=2 \text { or }\left(2^{x}-2\right)\left(2^{x}+3\right)=0 \\
x=2, \frac{1}{2} \text { or } 1
\end{array}
$$

Sum of all real roots $=3.5$