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

Question

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?

Answer 1

The following procedure, I believe, is somewhat simpler

$$\begin{equation}\begin{aligned} \left(x+{\frac{1}{x}}\right)^4 \left(x-{\frac{1}{x}}\right)^2 & = \left(x+{\frac{1}{x}}\right)^2\left(x+{\frac{1}{x}}\right)^2\left(x-{\frac{1}{x}}\right)^2 \\ & = \left(x+{\frac{1}{x}}\right)^2\left(\left(x+{\frac{1}{x}}\right)\left(x-{\frac{1}{x}}\right)\right)^2 \\ & = \left(x+{\frac{1}{x}}\right)^2\left(x^2 - \frac{1}{x^2}\right)^2 \\ & = \left(\left(x+{\frac{1}{x}}\right)\left(x^2 - \frac{1}{x^2}\right)\right)^2 \\ & = \left(x^3 - \frac{1}{x} + x - \frac{1}{x^3}\right)^2 \\ & = \left(\left(x^3 - \frac{1}{x^3}\right) + \left(x - \frac{1}{x}\right)\right)^2 \\ & = \left(x^3 - \frac{1}{x^3}\right)^2 + 2\left(x^3 - \frac{1}{x^3}\right)\left(x - \frac{1}{x}\right) + \left(x - \frac{1}{x}\right)^2 \\ & = x^6 - 2 + \frac{1}{x^6} + 2\left(x^4 - x^2 - \frac{1}{x^2} + \frac{1}{x^4}\right) + x^2 - 2 + \frac{1}{x^2} \\ & = x^6 + 2x^4 - x^2 - 4 -\frac{1}{x^2} + \frac{2}{x^4} + \frac{1}{x^6} \end{aligned}\end{equation}\tag{1}\label{eq1}$$

Answer 2

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).

Answer 3

$$(x+{\frac{1}{x}})^4 (x-{\frac{1}{x}})^2= \frac {(x^2+1)^4(x^2-1)^2}{x^6}=$$

$$\frac {(x^4-1)^2(x^2+1)^2}{x^6} =\frac {(x^8-2x^4+1)(x^4+2x^2+1)}{x^6}=$$

$$\frac{x^{12} +2x^{10} -x^8-4x^6-x^4+2x^2+1}{x^6}=$$

$$x^{6} +2x^{4} -x^2-4-\frac {1}{x^2}+\frac {2}{x^4}+\frac {1}{x^6}$$

Answer 4

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.

$(x + \frac 1x)^6 = (x^6 + \frac1{x^6}) + 6 (x^4 + \frac1{x^4}) + 15 (x^2 + \frac1{x^2}) + 20\\ (x + \frac 1x)^4 = (x^4 + \frac1{x^4}) + 4 (x^2 + \frac1{x^2}) + 6$

$(x + \frac 1x)^6 - 4(x + \frac 1x)^4 = (x^6 + \frac1{x^6}) + 2 (x^4 + \frac1{x^4}) - (x^2 + \frac1{x^2}) - 4$