differential equation $f(x)= x^3 +x^2 \cdot f (1) + x \cdot f'' (2)+f''' (3)$ If $f:\mathbb R\to \mathbb R$ such that $f(x)=x^3+x^2f'(1)+xf''(2)+f'''(3)$  ∀$x∈\mathbb R$ then find the value of $f(1)-f(0)$.
I am solving in this manner and I was stuck in $4$th  step. Please help me to continue to solve it.
$f(x)=x^3+x^2f'(1)+xf''(2)+f'''(3)$
Let ,  $f'(1)=\rho$ , $f''(2)=\theta$ and $f'''(3)=μ$
then
$$\begin{align}
f(x)&=x^3+\rho x^2+ \theta x+ \mu \\ 
f'(x)&=3x^2+2\rho x+\theta
\end{align}$$

Let $f(x)=x^3+ax^2+bx+c$ .
$$f'(x)=3x^2+2ax+b$$ so choosing $x=1$ :
$$a=3+2a+b$$
$$f''(x)=6x+2a$$ so choosing $x=2$ :
$$b=12+2a$$
$$f'''(x)=6$$ so $c=6$ .
Solving the system in $a$ and $b$ gives $a=-5$ and $b=2$ .
The function is thus : $$f(x)=x^3-5x^2+2x+6$$ and now simply find : $$f(1)-f(0)=4-6=-2$$
