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
 A: Here is how you advance. Try to simplify the expression as
$$(x+1/x)^3(x-1/x)^5= -\frac{1}{x^8}(1+x^2)^3(1-x^2)^5= -\frac{1}{x^8}(1-x^4)^3(1-x^2)^2$$
$$ = -\frac{1}{x^8}(1-x^2)^2(1-x^4)^3$$
$$ = -\frac{1}{x^8}(1-2x^2+x^4)(1-3x^{4}+3x^8-x^{12})=\dots\,.  $$
Can you finish it now?
A: I think you may as well simply expand the two binomials.  It's convenient to write the "cubic" from lowest to highest and the "quintic" from highest to lowest, so that we have
$$\left({1\over x}+x \right)^3\left(x-{1\over x} \right)^5=\left({1\over x^3}+{3\over x}+3x+x^3\right)\left(x^5-5x^3+10x-{10\over x}+{5\over x^3}-{1\over x^5} \right)$$
The coefficient for $x^2$ is thus
$$(1\cdot1)-(3\cdot5)+(3\cdot10)-(1\cdot10)=6$$
A: 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 $1(-1)+3\cdot5+3(-10)+1\cdot10=\cdots$
A: After applying my hint you are looking for the coefficient of $x^5$ in $(1+x)^3(-1+x)^5$. You can compute this directly, or with a little less effort by recognising $(1+x)(-1+x)=(-1+x^2)$ so that the expression becomes $(-1+x^2)^3(-1+x)^2$ (cf. the answer by Mhenni Benghorbal); computing the coefficient of $x^5$ there is very easy since only the odd exponent in $(-1+x)^2$ contributes, and gives$~-3\times-2=+6$.
