Giving a formula for the coefficient of $x^k$ in the expansion of $(x^2 − 1/x)^{100}$ , where $k$ is an integer I followed the binomial theorem and got this:
The binomial theorem is:$$
(a+b)^n=\sum^n_{k=0}\binom{n}{k}a^kb^{n−k}.
$$
Then let $a=x$, $b=\dfrac1x$, $n=n$, $k=k$, I get$$
\sum^n_{j=0}\binom{n}{j}x^j\left(\frac1x\right)^{n−j}.
$$
I am not sure if I am doing it right. Can someone help me out with this?
 A: Note that $$\left(x^2-\frac1x\right)^{100}=\frac1{x^{100}}(x^3-1)^{100}$$ so use the Binomial Theorem from here.
A: Using binomial theorem:
$$\left(x^2 - \frac{1}{x}\right)^{100} = \sum_{k=0}^{100} {100 \choose k} x^{2k} (-1)^{100-k} \frac{1}{x^{100-k}} = \sum_{k=0}^{100} (-1)^{100-k} {100 \choose k} x^{3k-100} $$ 
Use that $3k-100 = m \in \mathbb{Z}$ if $k=\frac{m+100}{3}$
So, $$\text{Coef } x^m = \left\lbrace \begin{array}{ll} (-1)^{\frac{200-m}{3}} {100 \choose \frac{m+100}{3}} & \text{ if } m+100 \text{ is multiple of 3}\\
0 & \text{ other case } \end{array}\right.$$
A: The question was about $\left(x^2 − \frac 1x\right)^{100},$ not 
$\left(x + \frac 1x\right)^{100},$ is that right?
If you want to set $(a+b)^{100} = \left(x^2 − \frac 1x\right)^{100},$
then you want $a = x^2$ and $b = -\frac 1x.$
You do not want to use $x$ (without squaring it) or $\frac1x$ (without the negative sign) as your terms.
If you put the correct $a$ and $b$ into your formula then you will be on the track to the correct answer.
