3. In the tetrahedron $ABCD$, it is known that a plane parallel to $AB$ and $CD$ intersects the edge $BD$ in the ratio $k$. Then the ratio of the volumes of the two parts formed by the plane cutting the tetrahedron is $\qquad$
3. $\frac{k^{3}+3 k^{2}}{3 k+1}$.

As shown in Figure 3, complete the tetrahedron into a triangular prism, and let the volume of tetrahedron $ABCD$ be $V$.

The plane $PMNS$ is parallel to $AB$ and $CD$, dividing the tetrahedron $ABCD$ into two parts. Let the volume of the part containing point $B$ be $V_{1}$, and the volume of the part containing point $D$ be $V_{2}$. Then $V_{1}+V_{2}=V$, the volume of the quadrilateral pyramid $A-CDFE$ is $2V$, and the volume of the triangular prism $AEF-BCD$ is $3V$.
Let $BP: PD=k$. Then $AR: AE=BP: BD=k:(k+1)$.
Thus, $V_{A-MNRQ}: V_{A-DCFE}=k^{3}:(k+1)^{3}$,
$V_{ARQ-BSP}: V_{AEF-BCD}=k^{2} ;(k+1)^{2}$.
Therefore, $V_{1}=V_{ARQ-RSP}-V_{A-MNRQ}=\frac{k^{3}+3 k^{2}}{(k+1)^{3}} V$,
$$
V_{2}=V-V_{1}=\frac{3 k+1}{(k+1)^{3}} V .
$$