$f(x) - f'(x) = x^3 + 3x^2 + 3x +1; f(9) =?$ Given the following $f(x) - f'(x) =  x^3 + 3x^2 + 3x +1$
Calculate $f(9) = ?$
I have tried to play with different number of derivatives. Also tried to solve it by equations.
Maybe there is some geometric meaning that could shade the light ?
I feel it is no complex problem at all.
Thanks for your solution

It is most convenient to search $f$ among the third degree polynomials with the leading coefficients $1$. Let $f_0 (x) = x^3 + a x^2 + b x + c$. Then the equation becomes $$(f_0 - f'_0) (x) := x^3 + (a - 3) x^2 + (b - 2a) x + (c - b) = x^3 + 3 x^2 + 3 x + 1.$$ Hence, we have $a = 6$, $b = 15$ and $c = 16$. Then $$f_0 (x) = x^3 + 6 x^2 + 15 x + 16.$$ But this is only one from a class of solutions. Now, let $f = f_0 + g$, where $(g - g') (x) \equiv 0$. This gives us $g (x) = c \exp x$ for any constant $c$. Thus, we have $$f (x) = x^3 + 6 x^2 + 15 x + 16 + c \exp x.$$ Hence, $f (9) = 1366 + c e^9$.
