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.
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 

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