6. As shown in Figure 2, let $A-BCD$ be a regular tetrahedron (the base $BCD$ is an equilateral triangle), and construct $AO \perp$ the base $BCD$, with $O$ being the foot of the perpendicular. $P$ is a point on the height $AO$ such that $PA = \frac{1}{m} AO (m > 1)$. A plane through point $P$ parallel to the base $BCD$ intersects the three edges $AB$, $AC$, and $AD$ at points $B_1$, $C_1$, and $D_1$, respectively. Point $Q$ lies on the segment $PO$, and a plane through point $Q$ parallel to the base $BCD$ bisects the volume of the frustum $BCD - B_1C_1D_1$. Then $\frac{PQ}{QO}$ equals ( ).
(A) $\frac{\sqrt[3]{2(m^3 + 1)} - 1}{\sqrt[3]{2(m^3 + 1)} - m}$
(B) $\frac{\sqrt[3]{m^3 + 1} - 1}{2m - \sqrt[3]{m^3 + 1}}$
(C) $\frac{\sqrt[3]{m^3 + 1} - 1}{\sqrt[3]{m^3 + 1} - m}$
(D) $\frac{\sqrt[3]{4(m^3 + 1)} - 2}{2m - \sqrt[3]{4(m^3 + 1)}}$
6.D.

Let $A O=a$, then
$$
A P=\frac{a}{m}, V_{A-B_{1}} c_{1} D_{1}=\frac{1}{m^{3}} V_{A-B C D} .
$$

Thus, $V_{B C D-B_{1} c_{1} D_{1}}=\left(1-\frac{1}{m^{3}}\right) V_{A-B C D}$.
Suppose a plane parallel to the base $B C D$ is drawn through point $Q$, intersecting the three edges $A B, A C, A D$ at points $B_{2}, C_{2}, D_{2}$, respectively, then
$$
\begin{array}{l}
V_{A-B} C_{2} D_{2}=\frac{1}{m^{3}} V_{A-B C D}+\frac{1}{2}\left(1-\frac{1}{m^{3}}\right) V_{A-B C D} \\
=\frac{m^{3}+1}{2 m^{3}} V_{A-B C D} .
\end{array}
$$

Therefore, $\left(\frac{A Q}{A O}\right)^{3}=\frac{m^{3}+1}{2 m^{3}}$, that is
$$
\begin{array}{l}
\frac{A Q}{A O}=\frac{1}{m} \sqrt[3]{\frac{m^{3}+1}{2}} . \\
\text { Hence } \frac{P Q}{Q O}=\frac{A Q-A P}{A O-A Q}=\frac{\frac{1}{2} \sqrt[3]{4\left(m^{3}+1\right)}-1}{m-\frac{1}{2} \sqrt[3]{4\left(m^{3}+1\right)}} \\
=\frac{\sqrt[3]{4\left(m^{3}+1\right)}-2}{2 m-\sqrt[3]{4\left(m^{3}+1\right)}}(m>1) .
\end{array}
$$