11 ・11 Let $(x+a)^{4}=x^{4}+a_{1} x^{3}+a_{2} x^{2}+a_{3} x+a_{4}$. If $a_{1}+a_{2}+a_{3}=64$, then $\alpha$ equals
(A) -4 .
(B) 4 .
(C) -2 .
(D) 2 .
(China Shaanxi Province Junior High School Mathematics Competition, 1997)
[Solution] Let $x=1$, then by the given condition,
$$
(1+\alpha)^{4}=1+a_{1}+a_{2}+a_{3}+a_{4}=65+a_{4} \text {. }
$$

Let $x=0$, then $\alpha^{4}=a_{4}$.
$$
\therefore \quad(1+\alpha)^{4}-\alpha^{4}=65 \text {. }
$$

From this, we get $2 \alpha+3 \alpha^{2}+2 \alpha^{3}-32=0$,
which is
$$
(\alpha-2)\left(2 \alpha^{2}+7 \alpha+16\right)=0 \text {. }
$$

And
$$
\begin{array}{l}
\text { and } \\
\therefore \quad \alpha=2 .
\end{array} \quad 2 \alpha^{2}+7 \alpha+16 \neq 0,
$$

Therefore, the answer is $(D)$.