Smirnov Oleg

A cube is given. Three planes, parallel to the faces, divide it into 8 parallelepipeds. They are painted in a checkerboard pattern. The volumes of the black parallelepipeds are $1, 6, 8, 12$.

Find the volumes of the white parallelepipeds.
Let's denote the intersection point of three planes cutting through a cube as . Note that the volume of any of the eight resulting parallelepipeds is equal to the product of the three edges emanating from . Suppose we want to find the volume of a certain white parallelepiped $\alpha$. By multiplying the volumes of the three black parallelepipeds adjacent to $\alpha$, we obtain the product of the lengths of nine segments: the edges of parallelepiped $\alpha$ emanating from  will appear in the product twice, while the edges of the black parallelepiped $\beta$ opposite to $\alpha$ and emanating from  will appear once. Dividing the resulting number by the volume of $\beta$, we get the square of the volume of $\alpha$.

Thus, by multiplying triples of numbers from the condition and dividing by the fourth, we obtain the squares of the volumes of the white parallelepipeds; it remains to extract the roots.

## Answer

$2,3,4,24$.