For $a \in \mathbb R$, find polynomials $P$ such that $(x^2-ax+18)P(x)-(x^2+3x)P(x-3)=0$ Find all polynomials $P(x)$ with real coefficients such that:-    $(x^2-ax+18)P(x)-(x^2+3x)P(x-3)=0$
This is $a$ creating some big problems for me. I don't know what to do.
I am not able to figure out anything because of that $a$. The best I can figure out is that I will find the roots of $P(x)$ because $a$ is not computable as there is no way of finding out the zeros of $P(x-3)$. If I would have been able to find the roots of $(x^2-ax+18)$ then I would have been able to figure out what to do.
If there would have been no $a$ I would have found of the roots of $P(x)$ like for example $\alpha, \beta$ then I would have written out $P(x)$ in the form of $(x-\alpha)(x-\beta)Q(x)$ for $Q(x)$ being any polynomial. Then I would have tried to calculate the answer.
Any help would be appreciated

the function is not a polynomial
$$ (x^2-ax+18)P(x) = (x^2+3x)P(x-3) $$
Put $x=-3$,
$$ (27+3a)P(-3) = (9-9)P(-6) $$ $ (27+3a) \cdot P(-3)=0$, $27+3a=0$, then $a= -9$
$$ (x²-(-9)x+18)P(x) = (x²+3x)P(x-3) $$
$$(x^2+9x+18)P(x) = (x^2+3x)P(x-3) $$
$$(x+3)(x+6)P(x) = x(x+3)P(x-3) $$
$$(x+6)P(x) = xP(x-3)$$
$$(x+6)P'(x)+ xP(x) = xP'(x)+P(x-3)$$
$$(x+6)P'(x)+ xP(x) = xP'(x)+ \frac{(x+6)}{x} P(x)$$
$$6P'(x) = \frac{(x+6)}{x} P(x)-xP(x) $$
$$6P'(x) = ( 1+ \frac{6}{x}-x)P(x) $$
$$ \frac{P'(x)}{P(x)} = \frac{1}{6}+\frac{1}{x}-\frac{x}{6} $$
$$ \int \frac{P'(x)}{P(x)} = \int ( \frac{1}{6}+\frac{1}{x}-\frac{x}{6} ) $$
$$\log(P(x))+c = \frac{x}{6}-\frac{x^2}{12}+\log(x) $$
$$\log(P(x)) = \frac{x}{6}-\frac{x^2}{12}+\log(x)-c $$
$$ P(x) = \exp ( \frac{x}{6}-\frac{x^2}{12}+\log(x)-c ) $$
$$P(x) = x \cdot \exp(  \frac{x}{6}-\frac{x^2}{12}-c) $$
