<s>[INST] 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.
 [/INST] 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.
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