3. A regular tetrahedron with a volume of 1 is placed inside a cube. The minimum volume of this cube is $\qquad$
As shown in the figure, when the four vertices of a regular tetrahedron are also the vertices of a cube, the two figures share the same circumscribed sphere, at which point the volume of the cube reaches its minimum value.
Let the edge lengths of the regular tetrahedron and the cube be $a, b$, respectively, then
$$
\begin{array}{l}
R=\frac{\sqrt{6}}{4} a=\frac{\sqrt{3}}{2} b \Rightarrow b=\frac{\sqrt{2}}{2} a . \\
\text { Also } V_{\text {tetrahedron }}=\frac{\sqrt{2}}{12} a^{3}=1 \Rightarrow a^{3}=6 \sqrt{2},
\end{array}
$$

Therefore, $V_{\text {cube }}=b^{3}=\frac{\sqrt{2}}{4} \cdot 6 \sqrt{2}=3$.