factor the expression x^6-1, using the following methods

Question

Factor the expression x^6-1, using the following methods (a)Factor the expression by treating it as the difference of two squares, (x^3)^2-1^2 (b) Factor the expression, treating it as the difference of two cubes, (x^2)^3-1^3

Answer 1

a)

x^6 - 1

= (x^3)^2 - 1

= (x^3 + 1)(x^3 - 1)

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

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

 

b)

x^6 - 1

= (x^2)^3 - 1

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

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

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

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

 

 

Answer 2

 Have to use identity

 

   X²  - Y²  = ( X - Y ) ( X +Y )!
   X³ - Y³  = ( X - Y ) ( X² + xy + y²  ) 

 

       ^{FIRST  :} (x ⁶  - 1 ) = ( x³ + 1 ) ( x³ + 1 ) = ( x + 1 ) ( x² - x + 1 ) ( x -1) ( x² + x + 1 )