Prove that there is no term independent of $x$ in the binomial expansion of $\left(x-\frac 1x\right)^{19}$

Solution 1:

Note that $\left(x-\frac 1x\right)^{19}=x^{-19}(x^2-1)^{19},$ so the question is the same as asking whether $(x^2-1)^{19}$ has a term $x^{19}$. But this polynomial is even, so it has no odd terms at all!

Solution 2:

If you want to do this without the binomial theorem, in an intuitive way, maybe you can argue combinatorially:

in the first place, forget about the minus sign. It is irrelevant.

When you multiply out the binomial you see that you may regard each term as 19 boxes, in each of which you put either $x$ or $1/x$. (The term itself is obtained by multiplying together all of the entries in the boxes).

Now it is easy to see that since 19 is odd, there will always be an instance of either $x$ or $1/x$ after canceling all the $x$, $1/x$ pairs.