Expansion $(x+{\frac{1}{x}})^4 (x-{\frac{1}{x}})^2$ 
Expand $(x+{\frac{1}{x}})^4 (x-{\frac{1}{x}})^2$

Right now, I am able to expand this expression by simplifying it to:
$\frac{(x^2+1)^4 (x^2-1)^2}{x^6}$
I used the formula $(a+b)^2$ and $(a-b)^2$ a bunch of times to arrive at the answer. But, is there any simpler/smarter way to do this?

Let $y=\frac1{x}$.  Then,
\begin{align}
(x+\frac1{x})^4(x-\frac1{x})^2&=(x+y)^4(x-y)^2\\
&=(x+y)^2(x+y)^2(x-y)^2\\
&=(x+y)^2(x^2-y^2)^2\\
&=\big((x+y)(x^2-y^2)\big)^2\\
&=(x^3+x^2y-xy^2-y^3)^2\\
&=(x^3+x-\frac1{x}-\frac1{x^3})^2\\
&=\bigg(\frac{x^6+x^4-x^2-1}{x^3}\bigg)^2\\
&=\frac{x^{12} + 2x^{10} - x^8 - 4 x^6 - x^4 + 2 x^2 + 1}{x^6}
\end{align}
(or $x^6+2x^4-x^2-4-\frac1{x^2}+\frac2{x^4}+\frac1{x^6}$, if you'd like).
