How do you factor #x^8 - y^8#?

1 Answer
Jan 23, 2018

#x^8-y^8 = (x-y)(x+y)(x^2+y^2)(x^2-sqrt(2)xy+y^2)(x^2+sqrt(2)xy+y^2)#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We can get some of the way by using this #3# times.

Then to factor the remaining quartic, we can use:

#(a^2-kab+b^2)(a^2+kab+b^2) = a^4+(2-k^2)a^2b^2+b^4#

with #k = sqrt(2)# to find:

#a^4+b^4 = (a^2-sqrt(2)ab+b^2)(a^2+sqrt(2)ab+b^2)#

So:

#x^8-y^8 = (x^4)^2-(y^4)^2#

#color(white)(x^8-y^8) = (x^4-y^4)(x^4+y^4)#

#color(white)(x^8-y^8) = ((x^2)^2-(y^2)^2)(x^4+y^4)#

#color(white)(x^8-y^8) = (x^2-y^2)(x^2+y^2)(x^4+y^4)#

#color(white)(x^8-y^8) = (x-y)(x+y)(x^2+y^2)(x^4+y^4)#

#color(white)(x^8-y^8) = (x-y)(x+y)(x^2+y^2)(x^2-sqrt(2)xy+y^2)(x^2+sqrt(2)xy+y^2)#

Footnote

Another way of finding the factorisation of #a^4+b^4# goes like this...

#a^4+b^4 = (a^2+b^2)^2 - 2a^2b^2#

#color(white)(a^4+b^4) = (a^2+b^2)^2 - (sqrt(2)ab)^2#

#color(white)(a^4+b^4) = ((a^2+b^2) - sqrt(2)ab)((a^2+b^2) + sqrt(2)ab)#

#color(white)(a^4+b^4) = (a^2-sqrt(2)ab+b^2)(a^2+sqrt(2)ab+b^2)#