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Solution
$$x^{8}-1$$ can be written as $$(x^{4}-1)(x^{4}+1)$$, which, in turn, can be written as $$(x^{2}-1)(x^{2}+1)(x^{4}+1)$$
$$x^{4}+2x^{3}-2x-1$$ can be written as $$(x^{2}-1)(x^{2}+2x +1)$$.
Hence, we can see that $$x^{2} - 1$$ is the HCF.
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