$8 \cdot 4$ satisfies the number of real $x$ that satisfy the equation $3^{2 x+2}-3^{x+3}-3^{x}+3=0$
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(E) 4.
(33rd American High School Mathematics Examination, 1982)
［Solution］Transform the equation

We have
$$
3^{2 x+2}-3^{x+3}-3^{x}+3=0 \text {, }
$$

which is $\square$
$$
3^{2}\left(3^{x}\right)^{2}-3^{x} \cdot\left(3^{3}+1\right)+3=0 \text {, }
$$
$9\left(3^{x}\right)^{2}-28\left(3^{x}\right)+3=0$,
or
$$
\left(3^{x}-3\right)\left[9\left(3^{x}\right)-1\right]=0 \text {. }
$$

Therefore, $3^{x}=3$, or $3^{x}=\frac{1}{9}$.
Thus, $\quad x=1$, or $x=-2$.
Hence, the answer is $(C)$.