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Here is a dissection based on Gerhard's latest idea. I didn't add this directly to his post because it is not identical to his $5$-piece dissection from unfoldings.

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  [![TetraCubeGood][1]][1]

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The tetrahedron is unfolded to a $3.722 \times 1.611 = 6$ rectangle. The cube is unfolded to two $3 \times 1$ rectangles, which are stacked to $3 \times 2$. This leaves a $0.388 \times 3 = 1.164$ rectangle to cover a $0.722 \times 1.611 = 1.164$ rectangle. This is accomplished by the magic of the Bolyai-Gerwien proof.

This appears to be $5$ pieces to form the $3.722 \times 1.611$ rectangle from the cube's two $3 \times 1$ rectangles. I consider this a solution, starting from surface unfoldings, and have accepted Gerhard's answer.

Here is a dissection based on Gerhard's latest idea. I didn't add this directly to his post because it is not identical to his $5$-piece dissection from unfoldings.

---

  [![TetraCubeGood][1]][1]

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The tetrahedron is unfolded to a $3.722 \times 1.611 = 6$ rectangle. The cube is unfolded to two $3 \times 1$ rectangles, which are stacked to $3 \times 2$. This leaves a $0.388 \times 3 = 1.164$ rectangle to cover a $0.722 \times 1.611 = 1.164$ rectangle. This is accomplished by the magic of the Bolyai-Gerwien proof.

This appears to be $5$ pieces to form the $3.722 \times 1.611$ rectangle from the cube's two $3 \times 1$ rectangles. I consider this a solution, starting from surface unfoldings, and have accepted Gerhard's answer.

Here is a dissection based on Gerhard's latest idea. I didn't add this directly to his post because it is not exactly what he had in mind.

---

  [![TetraCubeGood][1]][1]

---

The tetrahedron is unfolded and reshaped to a $3.722 \times 1.611 = 6$ rectangle. The cube is unfolded to two $3 \times 1$ rectangles, which are stacked to $3 \times 2$. This leaves a $0.388 \times 3 = 1.164$ rectangle to cover a $0.722 \times 1.611 = 1.164$ rectangle. This is accomplished by the magic of the Bolyai-Gerwien proof.

This appears to be $5$ pieces to form the $3.722 \times 1.611$ rectangle from the cube's two $3 \times 1$ rectangles, and two more pieces to cut out the right triangle to reshape the rectangle back to the tetrahedron parallelogram. $7$ pieces altogether. I consider this a solution, and have accepted Gerhard's answer.

Here is a dissection based on Gerhard's latest idea. I didn't add this directly to his post because it is not identical to his $5$-piece dissection from unfoldings.

---

  [![TetraCubeGood][1]][1]

---

The tetrahedron is unfolded to a $3.722 \times 1.611 = 6$ rectangle. The cube is unfolded to two $3 \times 1$ rectangles, which are stacked to $3 \times 2$. This leaves a $0.388 \times 3 = 1.164$ rectangle to cover a $0.722 \times 1.611 = 1.164$ rectangle. This is accomplished by the magic of the Bolyai-Gerwien proof.

This appears to be $5$ pieces to form the $3.722 \times 1.611$ rectangle from the cube's two $3 \times 1$ rectangles. I consider this a solution, starting from surface unfoldings, and have accepted Gerhard's answer.

Here is a dissection based on Gerhard's latest idea. I don't add this directly to his post because I am not certain it is exactly what he had in mind.

---

  [![TetraCubeGood][1]][1]

---

The tetrahedron is unfolded and reshaped to a $3.722 \times 1.611 = 6$ rectangle. The cube is unfolded to two $3 \times 1$ rectangles, which are stacked to $3 \times 2$. This leaves a $0.388 \times 3 = 1.164$ rectangle to cover a $0.722 \times 1.611 = 1.164$ rectangle. This is accomplished by the magic of the Bolyai-Gerwien proof.

This appears to be $5$ pieces to form the $3.722 \times 1.611$ rectangle from the cube, and two more pieces to cut out the right triangle to reshape the rectangle back to the tetrahedron parallelogram. $7$ altogether.

Here is a dissection based on Gerhard's latest idea. I didn't add this directly to his post because it is not exactly what he had in mind.

---

  [![TetraCubeGood][1]][1]

---

The tetrahedron is unfolded and reshaped to a $3.722 \times 1.611 = 6$ rectangle. The cube is unfolded to two $3 \times 1$ rectangles, which are stacked to $3 \times 2$. This leaves a $0.388 \times 3 = 1.164$ rectangle to cover a $0.722 \times 1.611 = 1.164$ rectangle. This is accomplished by the magic of the Bolyai-Gerwien proof.

This appears to be $5$ pieces to form the $3.722 \times 1.611$ rectangle from the cube's two $3 \times 1$ rectangles, and two more pieces to cut out the right triangle to reshape the rectangle back to the tetrahedron parallelogram. $7$ pieces altogether. I consider this a solution, and have accepted Gerhard's answer.