How to show x and y are equal? I'm working on a proof to show that f: $\mathbb{R} \to \mathbb{R}$ for an $f$ defined as $f(x) = x^3 - 6x^2 + 12x - 7$ is injective. Here is the general outline of the proof as I have it right now:
Proof: For a function to be injective, whenever $x,y \in A$ and $x\neq y$, then $f(x) \neq f(y)$, i.e., where $A$, $B$ are finite sets, every two elements of $A$ must have distinct images in $B$, which also implies that there must be at least as many elements in $B$ as in $A$ such that the cardinality of $A$ is less than or equals the cardinality of $B$.
We shall prove the contra-positive: If $\exists$ $f(x) = f(y)$, then $x=y.$
Let $x^3 - 6x^2 + 12x - 7 = y^3 - 6y^2 + 12y - 7$.
Then by addition and some algebra, we get $x(x^2 - 6x + 12) = y(y^2 - 6y + 12)$

This feels dumb to ask but how do I continue to finally get the result that $x = y$?

A differentiable function with everywhere positive derivative is injective. The derivative of $f$ is $f'(x)=3x^2-12x+12=3(x^2-4x+4)=3(x-2)^2$. This is positive for $x\ne2$. To see that the zero derivative at $x=2$ doesn't destroy injectivity, integrate this to find $f(x)=(x-2)^3+C$ (with $C=1$). Thus $f$ is a shifted version of $x^3$, which is injective.
