<s>[INST] Packing of rectangular blocks There are 3 cubes side length of each a, b and c. There are also 3 identical rectangular blocks with sides measuring (a + b ), (b + c) and (c + a ).
Can they be assembled together to form a cube of side  (a + b + c ) ? ...
by virtue of the identity (a+b+c)^3 = a^3 + b ^3 + c^3 + 3( a + b )(b +c)( c+a).
If so how, if not, why not?
 [/INST] No.
The point is that if it could be done, faces of the whole cube would be composed of the faces of original blocks. Every block can contribute maximum of half of it's faces for the cubes faces (no two opposite faces can lie on the cubes faces), so the total area of the faces of the cube is at most the half of the sum of the areas of faces of the blocks.
The former is
$$
6(a+b+c)^2 = 6a^2+6b^2+6c^2+12ab+12bc+12ca
$$
and the latter
$$
3a^2+3b^2+3c^2+3((a+b)(b+c)+(b+c)(c+a)+(c+a)(a+b)) \\
= 6a^2+6b^2+6c^2+9ab+9bc+9ca.
$$
So, basically, the bricks don't have enough area for the cube.
</s>