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: 
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 
$$\frac{f^{(2)}(-c)}{2!} \stackrel{!}{=} -3 \Leftrightarrow \frac{6(-c)-24}{2!}=-3 \Leftrightarrow \boxed{c=-3}$$
You want a factorization of the form $u^2(u-3)$. This means, the given polynomial should also factor into $(x-c)^2(x-c-3)$.

So, the first thing to do, whether you can factor the given polynomial.

Based on the [rational root test](https://en.wikipedia.org/wiki/Rational_root_theorem) you find quickly that your polynomial factors into 

$$(x-3)^2(x-6) \Rightarrow u = x-3 \Leftrightarrow x= u+3$$