Question 1160863
<pre>


{{{(x+1/x)^(5)*(x-1/x)^(3)}}}

{{{((x^2+1)/x^"")^5((x^2-1)/x^"")^3}}}

{{{expr((x^2+1)^5/x^5)*expr((x^2-1)^3/x^3)}}}

{{{((x^2+1)^5(x^2-1)^3)/x^8}}}

Using Pascal's triangle:

{{{((x^10+5x^8+10x^6+10x^4+5x^2+1)(x^6-3x^4+3x^2-1))/x^8}}}

So we only need to multiply the terms which will give us
a term in x<sup>12</sup> on top since we have x<sup>8</sup> on the bottom, and
the exponents will subtract to give exponent 4.

{{{(x^10)(3x^2)+(5x^8)(-3x^4)+(10x^6)(x^6)}}}

{{{3x^12-15x^12+10x^12}}}

{{{-2x^12}}}

So the coefficient of x<sup>4</sup> in the original problem is also -2

Edwin</pre>