How do you factor completely #x^6 - y^6#?

Question

26387 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-21T09:47:52

Using the identity #a^2-b^2=(a+b)*(a-b)# we have that

#x^6-y^6=(x^3)^2-(y^3)^2=(x^3-y^3)*(x^3+y^3)#

Now we know that

#x^3-y^3=(x-y)*(x^2+xy+y^2)#

and

#x^3+y^3=(x+y)*(x^2-xy+y^2)#

So finally is

#x^6-y^6=(x-y)*(x+y)*(x^2-xy+y^2)*(x^2+xy+y^2)#

Answer 2 · 2015-11-21T09:50:36

#x^6-y^6 = (x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)#

The difference of squares identity can be written:

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

The difference of cubes identity can be written:

#a^3-b^3 = (a-b)(a^2+ab+b^2)#

The sum of cubes identity can be written:

#a^3+b^3 = (a+b)(a^2-ab+b^2)#

So:

#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)#

If we allow Complex coefficients, then this reduces into linear factors:

#=(x-y)(x-omega y)(x-omega^2 y)(x+y)(x+omega y)(x+omega^2 y)#

where #omega = -1/2+sqrt(3)/2 i = cos((2pi)/3) + sin((2pi)/3)i# is the primitive Complex cube root of #1#.