Question

factorise:x^8-1 plss and .fastt​

Answer 1

${x}^{8} - 1 \\ \\ = {( {x}^{4} )}^{2} - 1 \\ \\ = ( {x}^{4} + 1)( {x}^{4} - 1) \\ \\ = ( {x}^{4} + 1)( {x}^{2} + 1)( {x}^{2} - 1) \\ \\ = ( {x}^{4} + 1)( {x}^{2} + 1)(x + 1)(x - 1) \\ \\ \\ \\ \\ formulae \\ \\ {x}^{2} - {y}^{2} = (x - y)(x + y)$

hope it helps

Answer 2

Step-by-step explanation:

x^8 - 1 = (x^4)^2-1^2 = (x^4-1)(x^4+1)

= {(x^2)^2 -1^2)}(x^4+1)

= (x^2 -1)(x^2 +1)(x^4+1)

= (x+1)(x-1)(x^2 +1)(x^4 +1)

Hope , it helps you dear .