<s>[INST] If $\,x-\frac 1 x=k, \, k$ being any integer,then $\,\,x^5-\frac {1}{x^5}=?$ I am stuck with the following problem which one of friends gave me :

If $\,x-\frac 1 x=k, \, k$ being any integer,then $\,\,x^5-\frac {1}{x^5}=?$
The options are $\,\,k^5+4k^3+4k, \,k^5+5k^3+6k,\,k^5+5k^3+5k,\,k^5+5k^3+4k $.

We see that $x-\frac 1 x=k \implies x=\frac{k \pm \sqrt{k^2+4}}{2}$. Now putting this value to $\,\,x^5-\frac {1}{x^2}$ makes the calculation complicated.
Can anyone help? Thanks and regards to all.
EDIT: The problem contained a typo and thanks to @noam for pointing that out. Now using binomial expansion of $x^5-\frac{1}{x^5}$, we see that option 3 is the correct choice.
 [/INST] HINT:
Applying the binomial expansion
$$
k^5=\left(x-\frac1x\right)^5-5\left(x^3-\frac1{x^3}\right)+10k.
$$
Now expand $k^3=\left(x-\frac1x\right)^3$ to express the cubic factor in terms of $k$.
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