<s>[INST] 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?
 [/INST] 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.
$(x + \frac 1x)^6 = (x^6 + \frac1{x^6}) + 6 (x^4 + \frac1{x^4}) + 15 (x^2 + \frac1{x^2}) + 20\\
(x + \frac 1x)^4 = (x^4 + \frac1{x^4}) + 4 (x^2 + \frac1{x^2}) + 6$
$(x + \frac 1x)^6 - 4(x + \frac 1x)^4 = (x^6 + \frac1{x^6}) + 2 (x^4 + \frac1{x^4}) - (x^2 + \frac1{x^2}) - 4$
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