The HCF of x^{8}-1 and x^{4}+2x^{3}-2x-1 is | vedclass

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(C) To find the $HCF$,we factorize both expressions.First expression: $x^{8}-1 = (x^{4}-1)(x^{4}+1) = (x^{2}-1)(x^{2}+1)(x^{4}+1) = (x-1)(x+1)(x^{2}+1

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$x^{2}-1$

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(C) To find the $HCF$,we factorize both expressions.First expression: $x^{8}-1 = (x^{4}-1)(x^{4}+1) = (x^{2}-1)(x^{2}+1)(x^{4}+1) = (x-1)(x+1)(x^{2}+1)(x^{4}+1)$.Second expression: $x^{4}+2x^{3}-2x-1 = (x^{4}-1) + 2x(x^{2}-1) = (x^{2}-1)(x^{2}+1) + 2x(x^{2}-1) = (x^{2}-1)(x^{2}+2x+1) = (x-1)(x+1)(x+1)^{2} = (x-1)(x+1)^{3}$.The common factors are $(x-1)$ and $(x+1)$.Therefore,the $HCF = (x-1)(x+1) = x^{2}-1$.

The $HCF$ of $x^{8}-1$ and $x^{4}+2x^{3}-2x-1$ is

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