<s>[INST] 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
 [/INST] $\dfrac{x^{2014}(x-1)}{(x-1)^3}= \dfrac{x^{2014}}{(x-1)^2}$;
$x^{2014}= Q(x)(x-1)^2+ ax+b$;
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 
$(x-1)(2014x-2013)$.
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