$6 \cdot 17$ If the base and exponent of $a^{b}$ are both tripled, the resulting number is denoted as $r$, where $a, b>0$. If $r$ equals the product of $a^{b}$ and $x^{b}$, where $x>0$, then $x$ equals
(A) 3.
(B) $3 a^{2}$.
(C) $27 a^{2}$.
(D) $2 a^{3 b}$.
(E) $3 a^{2 b}$.
(44th American High School Mathematics Examination, 1993)
[Solution] Since $r=(3 a)^{3 b}=\left[(3 a)^{3}\right]^{b}=\left(27 a^{3}\right)^{b}$, and $r=a^{b} x^{b}=(a x)^{b}$,
we have $\left(27 a^{3}\right)^{b}=(a x)^{b}$, which means $27 a^{3}=a x$, thus $x=27 a^{2}$. Therefore, the answer is $(C)$.