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: 
Here's a relatively fast approach that avoids factoring a cubic.

Notice that $u^3 - 3 u^2 = u^2 (u - 3)$ has a double root at $u = 0$, so if $x = u + c$, the given polynomial, $$p(x) := x^3 − 12 x^2 + 45 x − 54$$ has a double root at $x = (0) + c = c$.

Since $c$ is a double root, it is also a root of $$p'(x) = 3 (x^2 - 8 x + 15) = 3 (x - 3) (x - 5) ,$$ so $c = 3$ or $c = 5$. But $5 \not\mid 54$, so the [Rational Root Theorem](https://en.wikipedia.org/wiki/Rational_root_theorem) implies that $p(5) \neq 0$, leaving
$$\color{#df0000}{\boxed{c = 3}} .$$
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 

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