Question: Find the value of $c$ so that the substitution $x=u+c$ transforms the equation $x^3-12x^2+45x-54=0$ into the equation $u^3-3u^2=0$

My first idea is to find the roots of $u^3-3u^2=0$ And getting $u=0$ and $u=3$, however I am not sure how to continue.

If someone could explain how to tackle this sort of problem, I would greatly appreciate it. Thanks.

Answer: 
$$f(u+c)=(u+c)^3-12(u+c)^2+45(u+c)-54$$
In the final equation the coefficient of $u$ is $0$.  

That means want to eliminate coefficient of $u$ in $f(u+c)$,  

so collect the coefficients of $u$ and set it to $0$:
$$\begin{align}&3c^2-12(2c) + 45=0\\&c^2-8c+15=0\\&c=3,5\end{align}$$

Plug each of these in $f(u+c)$ and see if the constant term also gets eliminated.
Another more systematic way without factorization uses Taylor polynomials.

Set $f(x) = x^3-12x^2+45x-54$. Then according Taylor you have around any $a \in \mathbb{R}$
$$f(x) = \sum\_{k=0}^{3}\frac{f^{(k)}(a)}{k!}(x-a)^k$$
$$\stackrel{a=-c}{\Longrightarrow}f(x) = \frac{f^{(2)}(-c)}{2!}(x+c)^2 + \frac{f^{(3)}(-c)}{3!}(x+c)^3$$

Since the coefficient in front of $u^2 = (x+c)^2$ is $-3$, the following equation must hold true 
$$\frac{f^{(2)}(-c)}{2!} \stackrel{!}{=} -3 \Leftrightarrow \frac{6(-c)-24}{2!}=-3 \Leftrightarrow \boxed{c=-3}$$