Here is a start:
As ihf observed, you need
$f(x)^3+g(x)^3+h(x)^3=x q(x)^3$
On the left side you have three polinomials of degrees multiple of 3, on the right the degree is M3+1, thus something must cancel on the left side.
Without loss of generality, you can consider 2 cases:
Case 1: $deg(f)=deg(g) = deg(h)$. In this case, let $ax^k, bx^k$ and $cx^k$ be the dominant monomials in $f,g,h$. Then
$a^n+b^n+c^n=0 \,.$ which is not possible by Fermat last Theorem. So this is not possible.
Case 2: $deg(f)=deg(g) \geq deg(h)+1$.
Then $f=ax^k+bx^{k-1}+...$. In order for the first two terms to cancel you need $g=-ax^k-bx^{k-1}+..$. And I am stucked here.
This suggest that the simplest potential example would be something like:
$f=x^2+ax+b \,;\, g=-x^2-ax+c \,;\, h(x)=dx+e \,.$
The computations seem too long :)