Problem 1345

Solution:
The answer is $(\mathbf{C})$. Expanding the cube, we get
$$(x+y+z)^{3}-x^{3}-y^{3}-z^{3}=6 x y z+3 x^{2} y+3 x y^{2}+3 x^{2} z+3 x z^{2}+3 y^{2} z+3 y z^{2}$$
$$=3 x y(x+y+z)+3 x z(x+y+z)+3 y z(y+z)$$
$$=3 x(x+y+z)(y+z)+3 y z(y+z)$$
$$=3(y+z)(x(x+y+z)+y z)$$
$$=3(y+z)(x(x+y)+z x+z y)$$
$$=3(y+z)(x(x+y)+z(x+y))$$
$$=3(y+z)(x+z)(x+y)$$

SECOND SOLUTION
The expression $(x+y+z)^{3}-x^{3}-y^{3}-z^{3}$ contains the monomial $6 x y z$. This rules out answers (A) and (D) because that monomial does not appear there, in answer (B) $18 x y z$ appears, in answer (E) $-9 x y z$ appears (moreover this polynomial is not homogeneous). The answer is therefore (C).

THIRD SOLUTION
The expression $(x+y+z)^{3}-x^{3}-y^{3}-z^{3}$ is made up of $3^{3}-3=24$ monomials (without combining like terms), all having coefficient 1. Of the 5 proposed expressions, (A), (B), (D) and (E) are made up of $3 \cdot 2 \cdot 3=18$ monomials (even if some are like terms) all having coefficient 1, only $(\mathbf{C})$ is made up of $3 \cdot 2^{3}=24$ monomials all having coefficient 1.