Question

factorisation: x⁸ - 1 ​

Answer 1

Hey there !

x^8-1

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

=(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)

Answer 2

Topic

Factorisation

To Factorise

$x^8 - 1$

Solution

Use formula, a² - b² = ( a + b )( a - b )

$x^8 - 1$

$(x^4)^2 - 1^2$

$( x^4 + 1 )( x^4 - 1 )$

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

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

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

Answer

So, answer is

$( x^4 + 1 )( x^2 + 1 )( x + 1 )( x - 1 )$.

Learn More :-

a³ - b³ = ( a - b )( a² + ab + b² )

a³ + b³ = ( a + b )( a² - ab + b² )

( a + b )² = a² + 2ab + b²

( a + b )² = a² + 2ab + b²