Question:
Factorise:
$x^{8}-1$
Solution:
$x^{8}-1$
$=\left(x^{4}\right)^{2}-1^{2}$
$=\left(x^{4}+1\right)\left(x^{4}-1\right) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$
$=\left(x^{4}+1\right)\left[\left(x^{2}\right)^{2}-1^{2}\right]$
$=\left(x^{4}+1\right)\left(x^{2}+1\right)\left(x^{2}-1\right) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$
$=\left(x^{4}+1\right)\left(x^{2}+1\right)\left(x^{2}-1\right)^{2}$
$=\left(x^{4}+1\right)\left(x^{2}+1\right)(x+1)(x-1) \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$