The following geometric proof of the Pythagoras theorem
relies on the fact that one can cut out 4 right angled triangles (of area $\frac12ab$ each) out of a square of side length $a+b$ to obtain another square. That is, $(a+b)^2-4\times\frac12ab$ is a square of a quantity, and it is easy to see that this quantity is the hypotenuse of the right-angled triangle of perpendicular sides $a$ and $b$.
In a similar vein, since $(a+b+c)^3 = a^3 + b^3 + c^3 + 3(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2) + 6abc$, is it possible to cut out finitely many tetrahedrons, whose total volume adds up to $3(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2) + 6abc$, out of a cube of edge-length $a+b+c$ and be left with a perfect cube with the volume of $a^3+b^3+c^3$?
In case it is possible, does this edge of length $(a^3+b^3+c^3)^{1/3}$ have a nice geometric interpretation?
Edit 1: The edge-lengths of each such tetrahedron should lie in $\mathbb{Q}[a,b,c]$, otherwise it is trivially possible.