Developing the binomial $[(x+\frac{1}{x})(x-\frac{1}{x})]^n$ we obtain independent term equal to $70$. The value $n$ is equal to?($S:n=8$)
$(x+\frac{1}{x})((x-\frac{1}{x})]^n=(x^2-\frac{1}{x^2} )^n\\ T_{p+1} = C(n,p) .x^{2{^({n-p)}}}.(\frac{-1}{x^2})^p =C(n,p) .x^{2n-2p-2p}.(-1)^p \implies 2n - 4p = 0$
$ \therefore n = 2p$
$T_{(\dfrac{n}{2}+1)} = C(2p,p).(x)^0 .(-1)^p \implies \dfrac{2p!}{p!.p!}.(-1)^p = 70$
How to finish?
You have ${2p \choose p}= \pm 70$. Central binomial coefficients as a function of $p$ are strictly increasing, and testing numbers gives 20 for $p=3$, $70$ for $p=4$ and $252$ for $p=5$. For $p<4$ it is too small in magnitude and $p>4$ are too big, while 4 itself works, so $p=4$, $n=2\cdot 4=8$ is the answer.