How to Factor (x^6)-(y^6) using difference of squares?

Question

anonymous · Answer

$$x^6-y^6$$

aum · Answer

$$
x^6 - y^6 = (x^3)^2 - (y^3)^2 = ?
$$

aum · Answer

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

anonymous · Answer

I got $$(x^3-y^3)(x^3+y^3) $$

but how would factor this even further?

anonymous · Answer

is that wrong?

aum · Answer

That is correct.  That is the first step.  Then you can factor further by using the following identities:
$$
a^3 - b^3 = (a-b)(a^2 + ab + b^2) \
a^3 + b^3 = (a+b)(a^2 - ab + b^2) 
$$

anonymous · Answer

oh okay the difference of cubes, thats whats i was missing i think,

anonymous · Answer

wait hold on, how would i go about plugging $$(x^3-y^3)(x^3+y^3)$$

anonymous · Answer

into the identity

freckles · Answer

replace a with x and replace y with b

anonymous · Answer

so I just do this:

$$(x-y)^3(x+y)^3=((x-y)-(x+y)((x-y)^2+(x-y)(x+y)+(x+y))$$

anonymous · Answer

and then simplify

freckles · Answer

didn't you have
$$(x^3-y^3)(x^3+y^3)?$$

anonymous · Answer

yes

freckles · Answer

use what aum gave you to factor more 
$$a^3 - b^3 = (a-b)(a^2 + ab + b^2) \ a^3 + b^3 = (a+b)(a^2 - ab + b^2)$$

freckles · Answer

like you can factor x^3-y^3 more by saying (x-y)(x^2+xy+y^2)
now you need to factor x^3+y^3

anonymous · Answer

Thanks guys I got it :D