Find the remainder of $\frac{x^{2015}-x^{2014}}{(x-1)^3}$ Find the remainder of $\frac{x^{2015}-x^{2014}}{(x-1)^3}$.
Let $P(x)=x^{2015}-x^{2014}=Q(x)(x-1)^3+ax^2+bx+c.$ If we put $x=1$ in $P(x)$ and $P'(x)$, we get $a+b+c=0$ and $2a+b=1$. Then: $c=a-1$. The second derivative won't help in finding $b$, so, what should I do? Thank you
 A: Rewrite expression as
$$
\frac{x^{2015} - x^{2014}}{(x-1)^3} = \frac{x^{2014}}{(x-1)^2}
$$
so we have
$$
x^{2014} = Q(x) (x-1)^2 + ax + b.
$$
Now you need to find coefficients $a$ and $b$. Your argument with derivatives should work:
$$
1^{2014} = 1 = a + b
$$
$$
2014\cdot 1^{2013} = 2014 = a.
$$
Hence, $b = -2013$ and the final answer is
$$
x^{2015} - x^{2014} = x^{2014} (x-1) = Q(x) (x-1)^3 + (x-1)\cdot(2014x - 2013)
$$
A: $\dfrac{x^{2014}(x-1)}{(x-1)^3}= \dfrac{x^{2014}}{(x-1)^2}$;
$x^{2014}= Q(x)(x-1)^2+ ax+b$;
Binomial expansion:
$x^{2014}=(1+(x-1))^{2014}=$ 
$\sum_{k=0}^{n}\binom{2014}{k}1^{n-k}(x-1)^k$
Remainder $ax+b$:
$\binom{2014}{0}1+\binom{2014}{1}(x-1)=$
$1+2014x-2014=2014x-2013$.
Originally: 
$x^{2014}(x-1)$ is divided by $(x-1)^3$:
Hence 
$x^{2014}(x-1)=Q(x)(x-1)^3+(x-1)(2014x-2013),$
with quadratic remainder 
$(x-1)(2014x-2013)$.
A: Using $\ zf\,\bmod zg\ = \,z\, (f \bmod g)\, =\,  $ mod Distributive Law to factor out $\,  z = x\!-\!1$
$\,\begin{align} z(z\!+\!1)^{\large n} \bmod z^{\large 3} &=\, z\,(\color{#c00}{(1+ z)^{\large n}}\bmod \color{#c00}{z^{\large 2}})\qquad\ \ \  \text{[OP is }\, n = 2104]\\[.4em]
&=\,z\,(\color{#c00}{1+nz})\ \ \text{ by } \color{#c00}{\text{Binomial or Taylor }}\text{Theorem}\\[.4em]
&=\ n\!-\!1 + (1\!-\!2n)\,x + n\, x^{\large 2}\ \ \ \  {\rm by}\,\ z=x\!-\!1
\end{align}$
A: Another way:
Set $x-1=y$
$x^r=(1+y)^r\equiv1+\binom r1y+\binom r2y^2\pmod{y^3}$
$x^m-x^n\equiv y(m-n)+y^2\left(\binom m2-\binom n2\right)\pmod{y^3}$
Replace $y$ with $x-1$
A: I can use the division algorithm of a polynomial by another polynomial: in particular set $A(x)=x^{2014}$ and $B(x)=x^2-2x+1$. Proceeding with successive division, I obtain as quotient $$Q(x)=x^{2012}+2x^{2011}+3x^{2010}+\cdots + 2012x+2013$$ The general term of this polynomial is $nx^{2013-n}$. When $n=2013$, I obtain as polynomial rest: $R(x)=-2012x+4026x-2013=2014x-2013$. Before starting this process I have cancel out $(x-1)$ from the numerator and denominator of the fraction, so: $$Q(x)=(x^{2012}+2x^{2011}+3x^{2010}+\cdots + 2012x+2013)(x-1)$$ and $$R(x)=(2014x-2013)(x-1)$$
A: It is:
$$\frac{x^{2015}-x^{2014}}{(x-1)^3}=\frac{(x-1+1)^{2015}-(x-1+1)^{2014}}{(x-1)^3}=\\
=\small{\frac{\left[S(x)+{2015\choose 2}(x-1)^2+{2015\choose 1}(x-1)+1\right]-\left[T(x)-{2014\choose 2}(x-1)^2-{2014\choose 1}(x-1)-1\right]}{(x-1)^3}}=\\
=\frac{S(x)-T(x)+2014(x-1)^2+(x-1)}{(x-1)^3}=Q(x)+\frac{(x-1)(2014x-2013)}{(x-1)^3}.$$
Note: It is more expanded and direct version of lab bhattacharjee's answer.
