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The following identity is a bit isolated in the arithmetics of natural integers $$3^3+4^3+5^3=6^3.$$ Let $K_6$ be a cube whose side has length $6$. We view it as the union of $216$ elementary unit cubes. We wish to cut it into $N$ connected components, each one being a union of elementary unit cubes, such that these components can be assembled so as to form three cubes of sizes $3,4$ and $5$. Of course, the latter are made simultaneously: a component may not be used in two cubes. There is a solution with $9$ pieces.

What is the minimal number $N$ of pieces into which to cut $K_6$ ?

About connectedness: a piece is connected if it is a union of elementary cubes whose centers are the nodes of a connected graph with arrows of unit length parallel to the coordinate axes.

Edit. Several comments ask for a reference for the $8$-pieces puzzle, mentionned at first in the question. Actually, $8$ was a mistake, as the solution I know constists of $9$ pieces. The only one that I have is the photograph in François's answer below. Yet it is not very informative, so let me give you additional informations (I manipulated the puzzle a couple weeks ago). There is a $2$-cube (middle) and a $3$-cube (right). At left, the $4$-cube is not complete, as two elementary cubes are missing at the end of an edge. Of course, one could not have both a $3$-cube and a $4$-cube in a $6$-cube. So you can imagine how the $3$-cube and the imperfect $4$-cube match (two possibilities). Other rather symmetric pieces are a $1\times1\times2$ (it fills the imperfect $4$-cube when you build the $3$-, $4$- and $5$-cubes) and a $1\times2\times3$. Two other pieces have only a planar symmetry, whereas the last one has no symmetry at all.

The following identity is a bit isolated in the arithmetics of natural integers $$3^3+4^3+5^3=6^3.$$ Let $K_6$ be a cube whose side has length $6$. We view it as the union of $216$ elementary unit cubes. We wish to cut it into $N$ connected components, each one being a union of elementary unit cubes, such that these components can be assembled so as to form three cubes of sizes $3,4$ and $5$. Of course, the latter are made simultaneously: a component may not be used in two cubes. There is a solution with $8$ 9$pieces. What is the minimal number$N$of pieces into which to cut$K_6$? About connectedness: a piece is connected if it is a union of elementary cubes whose centers are the nodes of a connected graph with arrows of unit length parallel to the coordinate axes. Edit. Several comments ask for a reference for the$8$-pieces puzzle, mentionned at first in the question. Actually,$8$was a mistake, as the solution I know constists of$9$pieces. The only one that I have is the photograph in François's answer below. Yet it is not very informative, so let me give you additional informations (I manipulated the puzzle a couple weeks ago). There is a$2$-cube (middle) and a$3$-cube (right). At left, the$4$-cube is not complete, as two elementary cubes are missing at the end of an edge. Of course, one could not have both a$3$-cube and a$4$-cube in a$6$-cube. So you can imagine how the$3$-cube and the imperfect$4$-cube match (two possibilities). Other rather symmetric pieces are a$1\times1\times2$(it fills the imperfect$4$-cube when you build the$3$-,$4$- and$5$-cubes) and a$1\times2\times3$. Two other pieces have only a planar symmetry, whereas the last one has no symmetry at all. link text 4 added 900 characters in body The following identity is a bit isolated in the arithmetics of natural integers $$3^3+4^3+5^3=6^3.$$ Let$K_6$be a cube whose side has length$6$. We view it as the union of$216$elementary unit cubes. We wish to cut it into$N$connected components, each one being a union of elementary unit cubes, such that these components can be assembled so as to form three cubes of sizes$3,4$and$5$. Of course, the latter are made simultaneously: a component may not be used in two cubes. There is a solution with$8$pieces. What is the minimal number$N$of pieces into which to cut$K_6$? About connectedness: a piece is connected if it is a union of elementary cubes whose centers are the nodes of a connected graph with arrows of unit length parallel to the coordinate axes. Edit. Several comments ask for a reference for the$8$-pieces puzzle. The only one that I have is the photograph below. Yet it is not very informative, so let me give you additional informations (I manipulated the puzzle a couple weeks ago). There is a$2$-cube (middle) and a$3$-cube (right). At left, the$4$-cube is not complete, as two elementary cubes are missing at the end of an edge. Of course, one could not have both a$3$-cube and a$4$-cube in a$6$-cube. So you can imagine how the$3$-cube and the imperfect$4$-cube match (two possibilities). Other rather symmetric pieces are a$1\times1\times2$(it fills the imperfect$4$-cube when you build the$3$-,$4$- and$5$-cubes) and a$1\times2\times3\$. Two other pieces have only a planar symmetry, whereas the last one has no symmetry at all.