Binomial coefficient question? I'm unsure how to do these types of questions, so any help would be great:

Find the coefficient of $x^2$ in the expansion of $(x+1/x)^3(x-1/x)^5$

Thanks

On simplification we get $$\frac{(1+x^2)^3}{x^3}\cdot\frac{(1-x^2)^5}{-x^5}$$ 
$$=-\frac{(1+x^2)^3(1-x^2)^5}{x^8}$$
The coefficient of $x^2$ in  $(x+1/x)^3(x-1/x)^5$ 
$=-$ the coefficient of $x^{10}$ in the Binomial expansion of $$(1+x^2)^3(1-x^2)^5$$
Now, $$(1+x^2)^3(1-x^2)^5$$
$$=(1+3x^2+3x^4+x^6)(1-\binom51x^2+\binom52(x^2)^2-\binom53(x^2)^3+\binom54(x^2)^4-x^{10})$$
$$=(1+3x^2+3x^4+x^6)(1-5x^2+10x^4-10x^6+5x^8-x^{10})$$
So, the coefficient of $x^{10}$ is $1(-1)+3\cdot5+3(-10)+1\cdot10=\cdots$
