22. Let $P M O$ be a triangle with $P M=2$ and $\angle P M O=120^{\circ}$. Let $B$ be a point on $P O$ such that $P M$ is perpendicular to $M B$, and suppose that $P M=B O$. The product of the lengths of the sides of the triangle can be expressed in the form $a+b \sqrt[3]{c}$, where $a, b, c$ are positive integers, and $c$ is minimized. Find $a+b+c$.
22. 28
Extend $P M$ to a point $C$ such that $P C \perp O C$. Since $\angle P M O=120^{\circ}, \angle C M O=60^{\circ}$ and $\angle C O M=30^{\circ}$. Let $P B=x$ and $M C=a$. Then $C O=a \sqrt{3}$ and $O M=2 a$. Moreover, $\triangle P M B$ and $\triangle P C O$ are similar triangles. Thus, we have
$$
\frac{2}{2+a}=\frac{x}{x+2}
$$
so $x=4 / a$.
Furthermore, by the Cosine Law on side $P O$ of $\triangle P M O$, we have
$$
(x+2)^{2}=4+4 a^{2}+2 a
$$

Plugging in $x=4 / a$ and expanding, we have
$$
\frac{16}{a^{2}}+\frac{16}{a}+4=4+4 a^{2}+4 a
$$
and so $4+4 a=a^{4}+a^{3}$. Hence $a^{3}=4$ and $a=\sqrt[3]{4}$. Thus, $x=4 / \sqrt[2]{4}$.
It follows that the product of the lengths of the sides of the triangle is
$$
(2 a)(2)(x+2)=(2 \sqrt[3]{4})(2)(2 \sqrt[3]{2}+2)=16+8 \sqrt[3]{4},
$$
so $a+b+c=16+8+4=28$.