To make the problem more easier to write down, suppose $x=\dfrac{1}{\sqrt[3]{2}}, y=1-x$.
Given seven cuboids with the following sizes:
(1) Three cuboids with size: $x\times x\times y$.
(2) Three cuboids with size: $x\times y\times y$.
(3) One cuboid with size: $y\times y\times y$.
My question is: Is it possible to cut these cuboids into some pieces and then form a cube by using all of the pieces?
My claim is it is possible, but I couldn't find a method. Please help!
Yes, it is possible, since the two sets of polyhedra both have Dehn invariant $0$; a result of Sydler (1965) shows that any two collections of polyhedra with the same Dehn invariant can be cut into polyhedral pieces and reassembled to form the other.
However, in this case you don't need the full power of Sydler's result; you can apply the simpler Wallace–Bolyai–Gerwien theorem that any two polygons with the same area can be cut into pieces and reassembled into each other.
In particular, any two rectangles can. If they are very skinny, we cut them in half and paste the halves together until their aspect ratios are within a factor of $2$ of each other, and then apply the following transformation:
What this means for our purposes is that if $ab = cd$, then an $a\times b\times e$ box is equidecomposable into a $c\times d \times e$ box.
So we proceed as follows:
Convert a $y\times y\times y$ box into an $x\times \frac yx \times y$ box, by applying our rectangle procedure with slices in the first and second dimensions.
Stack this on top of the three $x\times y\times y$ boxes and the three $x\times x \times y$ boxes along their second dimension to get an $x\times (3x + 3y+\frac yx) \times y$ box.
Apply the rectangle-reassembly procedure to this box in the second and third dimensions to obtain an $x\times x \times x$ cube, and we're done. In fact, the procedure involved only translations of pieces, with no rotation needed.