Factorise:
Question:

Factorise:

$x^{8}-1$

 

Solution:

$x^{8}-1$

$=\left(x^{4}\right)^{2}-1^{2}$

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

$=\left(x^{4}+1\right)\left[\left(x^{2}\right)^{2}-1^{2}\right]$

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

$=\left(x^{4}+1\right)\left(x^{2}+1\right)\left(x^{2}-1\right)^{2}$

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

 

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