Take a cyclic polynomial and define its "symmetric part" as $\frac{P(x, y, z)+P(y, x, z)}{2}$ and its "antisymmetric part" as $\frac{P(x, y, z)-P(y, x, z)}{2}$. We can see that the symmetric part is in fact symmetric: switching the variables around any way leaves the sum invariant because $P(x, y, z)$ is cyclic. The antisymmetric part, when $x=y$, is 0. So $(x-y)$ is a factor of it. So is $(y-z)$ and $(z-x)$ for the same reason. We factor these out: $\frac{P(x, y, z)-P(y, x, z)}{2}=(x-y)(y-z)(z-x)R(x, y, z)$. Then switching $x$ and $y$ gives $(y-x)(x-z)(z-y)R(y, x, z)=\frac{P(y, x, z)-P(x, y, z)}{2}=-(x-y)(y-z)(z-x)R(x, y, z)$ so $R(x, y, z)=R(y, x, z)$. Clearly since $P$ is cyclic, so is $R$. Hence $R$ is symmetric too. By the fundamental theorem on symmetric polynomials, any symmetric polynomial of degree 3 can be written as a polynomial in $x+y+z$, $xy+yz+zx$, and $xyz$. So then if we let $P_1(x, y, z)=x+y+z$, $P_2(x, y, z)=xy+yz+zx$ and $P_3(x, y, z)=xyz$, then we can let $P_4(x, y, z)=(x-y)(y-z)(z-x)$ and by what we just proved, for any cyclic $P(x, y, z)$ there is a $Q$ as required in the problem.