1.91 Let the real number $p=\sqrt[3]{4}-\sqrt[3]{6}+\sqrt[3]{9}$, then $p$ satisfies
(A) $0<p<1$.
(B) $1<p<2$.
(C) $2<p<3$.
(D) $p=\sqrt[3]{7}$.
(1st "Five Sheep Cup" Junior High School Mathematics Competition, 1989)
And
$$
\frac{5}{\sqrt[3]{2}+\sqrt[3]{3}}>\frac{5}{2+3}=1 \text {. }
$$

Also, $1 \cdot 2^{3}<2,1 \cdot 3^{3}<3$,
thus
$$
\frac{5}{\sqrt[3]{2}+\sqrt[3]{3}}<\frac{5}{1.2+1.3}=2 \text {. }
$$

Therefore,
$$
1<p<2 \text {. }
$$

Hence, the answer is $(B)$.