FACTORING COMPLETELY

[gemini77]

factoring x^6 - y^6

difference of squares is (x-y)(x+y)

X^3-y^3 is (x-y)(x^2 + xy + y^2)

to ^4 is (x-y)(x+y) (x^2 + xy + y^2)

what is to the ^5 or ^6 look like??????

 

[tutor_joel]

you have the right idea.

$\displaystyle x^6-y^6 = (x^3)^2-(y^3)^2$

there's your difference of two squares. Then you have difference and sum of two cubes. Or...

$\displaystyle x^6-y^6 = (x^2)^3-(y^2)^3$

 

[masters]

factoring x^6 - y^6

difference of squares is (x-y)(x+y)

X^3-y^3 is (x-y)(x^2 + xy + y^2)

to ^4 is (x-y)(x+y) (x^2 + xy + y^2)

what is to the ^5 or ^6 look like??????

Hi gemini,

First, factor as the difference of two squares. Then, factor the sum and difference of two cubes.

$\displaystyle x^6-y^6=(x^3)^2-(y^3)^2=(x^3+y^3)(x^3-y^3)=(x+y)(x^2-xy+y^2)(x-y)(x^2+xy+y^2)$