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?
 A: The following procedure, I believe, is somewhat simpler
$$\begin{equation}\begin{aligned}
\left(x+{\frac{1}{x}}\right)^4 \left(x-{\frac{1}{x}}\right)^2 & = \left(x+{\frac{1}{x}}\right)^2\left(x+{\frac{1}{x}}\right)^2\left(x-{\frac{1}{x}}\right)^2 \\
& = \left(x+{\frac{1}{x}}\right)^2\left(\left(x+{\frac{1}{x}}\right)\left(x-{\frac{1}{x}}\right)\right)^2 \\
& = \left(x+{\frac{1}{x}}\right)^2\left(x^2 - \frac{1}{x^2}\right)^2 \\
& = \left(\left(x+{\frac{1}{x}}\right)\left(x^2 - \frac{1}{x^2}\right)\right)^2 \\
& = \left(x^3 - \frac{1}{x} + x - \frac{1}{x^3}\right)^2 \\
& = \left(\left(x^3 - \frac{1}{x^3}\right) + \left(x - \frac{1}{x}\right)\right)^2 \\
& = \left(x^3 - \frac{1}{x^3}\right)^2 + 2\left(x^3 - \frac{1}{x^3}\right)\left(x - \frac{1}{x}\right) + \left(x - \frac{1}{x}\right)^2 \\
& = x^6 - 2 + \frac{1}{x^6} + 2\left(x^4 - x^2 - \frac{1}{x^2} + \frac{1}{x^4}\right) + x^2 - 2 + \frac{1}{x^2} \\
& = x^6 + 2x^4 - x^2 - 4 -\frac{1}{x^2} + \frac{2}{x^4} + \frac{1}{x^6}
\end{aligned}\end{equation}\tag{1}\label{eq1}$$
A: $$(x+{\frac{1}{x}})^4 (x-{\frac{1}{x}})^2= \frac {(x^2+1)^4(x^2-1)^2}{x^6}=$$
$$\frac {(x^4-1)^2(x^2+1)^2}{x^6} =\frac {(x^8-2x^4+1)(x^4+2x^2+1)}{x^6}=$$
$$\frac{x^{12} +2x^{10} -x^8-4x^6-x^4+2x^2+1}{x^6}=$$
$$x^{6} +2x^{4} -x^2-4-\frac {1}{x^2}+\frac {2}{x^4}+\frac {1}{x^6}$$
A: 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).
A: Observe:
$(x - \frac{1}{x})^2 = x^2 - 2 + \frac {1}{x^2}\\
(x + \frac{1}{x})^2 = x^2 + 2 + \frac {1}{x^2}\\
(x - \frac{1}{x})^2= (x + \frac 1x)^2 - 4$
We make this substitution in the original expression
$(x + \frac 1x)^4(x - \frac{1}{x})^2)\\
(x + \frac 1x)^4((x + \frac{1}{x})^2 - 4)\\
(x + \frac 1x)^6 - 4(x + \frac{1}{x})^4$
Note, we are doing something not entirely unlike $\sin^2 x = 1 - \cos^2 x$ here.
And use the binomial expansion on each term above.
$(x + \frac 1x)^6 = (x^6 + \frac1{x^6}) + 6 (x^4 + \frac1{x^4}) + 15 (x^2 + \frac1{x^2}) + 20\\
(x + \frac 1x)^4 = (x^4 + \frac1{x^4}) + 4 (x^2 + \frac1{x^2}) + 6$
$(x + \frac 1x)^6 - 4(x + \frac 1x)^4 = (x^6 + \frac1{x^6}) + 2 (x^4 + \frac1{x^4}) - (x^2 + \frac1{x^2}) - 4$
