Find the remainder when $(x+1)^n$ is divided by $(x-1)^3$. I know this question has been answered before, but I have a slightly different different question.
I saw the solution of this question in my book and the author has solved it by substituting $x-1=y$ and then equating the coefficients of $y^2$, $y^1$ and $y^0$  to $A(y+1)^2$, $B(y+1)^1$ and $C$.
My question is why do we have to substitute $x-1=y$ and why can't equate coefficients of $x^2$, $x^1$ and $x^0$ to $A$, $B$ and $C$ without substituting? Thanks in advance.

Write $x:=u+1$. Then
$$(x+1)^n=(u+2)^n=\sum_{k=0}^n{n\choose k}2^{n-k}u^k\ .$$
Dividing by $(x-1)^3=u^3$ we get the remainder
$$r={n\choose0}2^n+{n\choose 1}2^{n-1}u+{n\choose 2}2^{n-2}u^2=2^n+n2^{n-1}(x-1)+n(n-1)2^{n-3}(x-1)^2\ .$$
This can be written in the form
$$r=2^{n-3}\bigl((n^2-n)x^2+(-2n^2+6n)x+(n^2-5n+8)\bigr)\ .$$
