Find the remainder when $(x+1)^n$ is divided by $(x-1)^3$ 
Find the remainder when $(x+1)^n$ is divided by $(x-1)^3$

I know that 
\begin{equation*}
(1 + x)^n = 1 + nx +\frac {n(n-1)}2!\cdot x^2 +\frac {n(n-1)(n-2)}3! \cdot x^3 +...    
\end{equation*}
but how can I use it to solve the above problem>Is there any other easier way to do it ?
 A: $$
(x+1)^n=((x-1)+2)^n=2^n+n 2^{n-1}(x-1)+\binom{n}{2}2^{n-2}(x-1)^2+(x-1)^3 \cdot F(x)
$$
So the remainder is $$2^n+n 2^{n-1}(x-1)+\binom{n}{2}2^{n-2}(x-1)^2.$$
A: You can write the equality (E):
$$(x+1)^n=P(x)(x-1)^3+a(x-1)^2+b(x-1)+c$$
where $P(x)$ is the quotient and $a(x-1)^2+b(x-1)+c$ the reminder of the euclidean division of $(x+1)^n$ by $(x-1)^3$. And your problem is to find $a,b,c$.
(E) can be seen as an equality between functions. So can you 
differentiate both side to get further equalities.
Make $x=1$ in the equality (E). You get $c=2^n$. Then differentiate the equality (E) one time and make again $x=1$ you get $b=n2^{n-1}$. Do it once again to get $a=\frac{n(n-1)}{2}2^{n-2}$.
Some background
I like very much Leox method as it is a way to find the reminder when you divide by $(x-1)^p$ with $p$ integer.
Do you also know the Taylor expansion of a polynomial $P(x)$ at a point $a$ ? Namely:
$$P(x)=P(a)+\frac{P^\prime(a)}{1!}(x-a)+\frac{P^{\prime\prime}(a)}{2!}(x-a)^2 + \dots + \frac{P^{(n)}(a)}{n!}(x-a)^n$$
Where $n$ is the degree of the polynomial $P$. This is the basic idea I used to provide the solution. This idea would work for all polynomial $P$.
A: By setting $y=x-1$, we just have to find the remainder of $(y+2)^n$ when divided by $y^3$.
By the binomial theorem:
$$(y+2)^n \equiv 2^n + n2^{n-1} y + n(n-1)2^{n-3} y^2\pmod{y^3}.$$
A: By the remainder theorem, the remainder when $p(x)$ is divided by $g(x)$, is $p(a)$ for $g(a)=0$. So in the above problem, $a=1$, therefore your answer should be $2^n$
