Functional dilemma

Let the polynomial have degree $d$ . The expression gives us:

$d=(d-1)+(d-2) \Rightarrow d=2d-3 \Rightarrow d=3$

So the polynomial is a cubic. Let $f(x)=a x^3+b x^2+c x+d$ and now equate co-efficients:

$f(2x)=f'(x) f''(x) \Rightarrow 8 a x^3+4 b x^2+2 c x+d=18 a^2 x^3+18 a b x^2+(6 a c +4 b^2) x+2 b c$

$8 a= 18 a^2 \Rightarrow a=\dfrac{4}{9} \: (a\neq 0)$

$4 b=18 a b \Rightarrow 4b=8b \Rightarrow b=0$

$2c=6ac+4b^2 \Rightarrow 2c=\dfrac{8c}{3}+0 \Rightarrow c=0$

$d=2bc \Rightarrow d=0\\ f(x)=\dfrac{4 x^3}{9}\\ f(3)= \dfrac{4 \times 3^3}{9}=\boxed{12}$

Let the degree of $f(x)$ be $D(f(x)) = n$ , then $D(f'(x)) = n-1$ and $D(f''(x)) = n-2$ . Then, we have:

$\begin{aligned} f(2x) & = f'(x)f''(x) \\ \implies D(f(2x)) & = D(f'(x)f''(x)) \\ n & = n-1 + n-2 \\ \implies n & = 3 \end{aligned}$

Therefore, $f(x)$ is of the form $f(x) = ax^3 + bx^2 + cx + d$ , where $a$ , $b$ , $c$ and $d$ are real numbers. Then, we have:

$\begin{aligned} f(2x) & = f'(x)f''(x) \\ 8ax^3 + 4bx^2 + 2cx + d & = (3ax^2 + 2bx + c)(6ax + 2b) \\ & = 18a^2x^3 + 18abx^2 + (2b^2+6ca)x + 2bc \end{aligned}$

Equating the coefficients:

$\begin{cases} 8a = 18a^2 & \implies a = \dfrac{4}{9} \\ 4b = 18ab = 8b & \implies b = 0 \\ 2c = 2b^2+6ca = 0 + \dfrac{8}{3}c & \implies c = 0 \\ d = 2bc & \implies d = 0 \end{cases}$

Therefore, $f(x) = \dfrac{4}{9}x^3 \implies f(3) = \dfrac{4}{9}(3^3) = \boxed{12}$