Question

divide (x raised to 8 -1) (x raised to 2 -1)

Answer 1

Given :

$\frac{x^8-1}{x^2-1}$

Important formula :

$(a+b)(a-b)=a^2-b^2$

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$\implies \frac{(x^4)^2-1^2}{x^2-1}$

$\implies \frac{(x^4+1)(x^4-1)}{x^2-1}$

$\implies \frac{(x^4+1)(x^2-1)(x^2+1)}{x^2-1}$

$\implies (x^4+1)(x^2+1)$

$\implies x^6+x^4+x^2+1$

Hope it helps you :-)

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