$8 \cdot 6$ satisfies $\left(2^{x}-4\right)^{3}+\left(4^{x}-2\right)^{3}=\left(4^{x}+2^{x}-6\right)^{3}$ for the sum of all real numbers $x$.
(A) $\frac{3}{2}$.
(B) 2 .
(C) $\frac{5}{2}$.
(D) 3 .
(E) $\frac{7}{2}$.
(42nd American High School Mathematics Examination, 1991)
[Solution] Let $2^{x}=u$, then $u>0$, and the given equation can be transformed into:

Then
$$
\begin{array}{c}
(u-4)^{3}+\left(u^{2}-2\right)^{3}=\left(u^{2}+u-6\right)^{3}, \\
\quad\left[(u-4)+\left(u^{2}-2\right)\right]\left[(u-4)^{2}-(u-4)\left(u^{2}-2\right)+\left(u^{2}-2\right)^{2}\right] \\
=\left(u^{2}+u-6\right)^{3},
\end{array}
$$

Rearranging terms, factoring out common factors, and simplifying, we get:
$$
\left(u^{2}+u-6\right)\left(u^{3}-4 u^{2}-2 u+8\right)=0,
$$

Then $u^{2}+u-6=0$, yielding $u=2$ (negative value discarded),
so $x=1$. And $u^{3}-4 u^{2}-2 u+8=0$,
which is $\left(u^{2}-2\right)(u-4)=0$, giving $u=4$ or $u^{2}=2$.
Thus, $2^{x}=4$, yielding $x=2$,
or $2^{x}=\sqrt{2}$, yielding $x=\frac{1}{2}$ (discarding $2^{x}=-\sqrt{2}$).
$\therefore$ The sum of the real roots of the equation is $2+1+\frac{1}{2}=\frac{7}{2}$.
Hence, the answer is $(E)$.