# galois theory – Does any cubic polynomial become reducible through composition with some quadratic?

What I mean to ask is this:

given an irreducible cubic polynomial $$P(X)in mathbb{Z}(X)$$ is there always a quadratic $$Q(X)in mathbb{Z}(X)$$ such that $$P(Q)$$ is reducible (as a polynomial, and then necessarily the product of 2 irreducible cubic polynomials)?

I did quite some testing and always found a $$Q$$ that does the job. For example:

$$P=aX^3+b,quad Q=-abX^2,quad P(Q)=-b(a^2bX^3-1)(a^2bX^3+1)$$

$$P=aX^3-x+1,quad Q=-aX^2+X,quad P(Q)=-(a^2X^3-2aX^2+X-1)(a^2X^3-aX^2+1)$$

and a particular hard one to find:

$$P=2X^3+X^2-X+4,quad Q=-8X^2+5X+1,quad P(Q)=(16X^3-18X^2+X+3)(64X^3-48X^2-11x-2)$$

Could there be a formula for $$Q$$ that works for all cases?

It feels to me that this may have a really basic Galois theoretic proof or explanation, but I can’t figure it out.