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.
 A: $$(x+1)^n=(x-1)^3Q (x)+ax^2+bx+c $$
for $x=1$, we get
$$\boxed {2^n=a+b+c} $$
differentiating, 
$$n (x+1)^{n-1}=3 (x-1)^2R (x )+2ax+b $$
with $x=1$,
$$\boxed {2^{n-1}=2a+b} $$
differentiating again
$$n (n-1)(x+1)^{n-2}=6 (x-1)S (x)+2a $$
with $x=1$, we find
$$\boxed {a=n (n-1)2^{n-3}}$$
A: $$
(x+1)^n = Q(x)(x-1)^3 + ax^2+b x+c
$$
$$ 2^n = a+b+c $$
$$
n(x+1)^{n-1} = Q'(x)(x-1)^3+3Q(x)(x-1)^2 + 2ax+b \Rightarrow n2^{n-1}=2a+b
$$
In the same way
$$
n(n-1)(n-2)2^{n-2} = 2a
$$
etc.
A: Let the quotient be $Q(x)$ and the remainder be $ax^2+bx+c$. Then
$$(x+1)^n=(x-1)^3Q(x)+ax^2+bx+c$$
If we put $y=x-1$, then we have
$$(y+2)^n=y^nQ(y+1)+a(y+1)^2+b(y+1)+c$$
We have
$$(y+2)^n=2^n+\binom{n}{1}2^{n-1}y+\binom{n}{2}2^{n-2}y^2+\textrm{terms involving higher powers of }y$$
So, we have 
$$a(y+1)^2+b(y+1)+c=2^n+\binom{n}{1}2^{n-1}y+\binom{n}{2}2^{n-2}y^2=2^n+2^{n-1}ny+2^{n-3}n(n-1)y^2$$
$a$, $b$ and $c$ can be found by comparing the coefficients.
A: 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
$$r=2^{n-3}\bigl((n^2-n)x^2+(-2n^2+6n)x+(n^2-5n+8)\bigr)\ .$$
A: 
My question is why do we have to substitute $x-1=y$

We do not have to use that substitution, but it makes calculations somewhat easier to follow. The idea is the same as writing $\,(x+1)^n=\big((x-1)+2\big)^n\,$ then retaining just the last three terms of the binomial expansion, since the rest have $\,(x-1)^3\,$ as a factor.

and why can't equate coefficients of $x^2$, $x^1$ and $x^0$ to $A$, $B$ and $C$ without substituting?

That doesn't work, because the Euclidean division is $\,(x+1)^n=(x-1)^3 q(x) + ax^2+bx+c\,$. It is straightforward to determine the coefficients of $\,x^0,x^1,x^2\,$ on the LHS, but not so easy on the other side since you don't know the quotient $\,q(x)\,$ which also contributes to the $\,x^0,x^1,x^2\,$ terms on the RHS, and calculating $\,q(x)\,$ would take some non-trivial (and not directly necessary) work.
