How do I factor the polynomial, x^6 - 64y^6?

Question

Please show all the steps, thanks

Answer 1

believe me, it's not difficult
x^6-64y^6= (x^3-8y^3)(x^3+8y^3)
=(x-2y)(x^2+xy+y^2)(x+2y)(x^2- xy+y^2)
That's all!
Hopefully, it'll help you:D

Answer 2

x^6 - 64y^6?
Notice that
x^6 = (x^3)^2
64y^6 = (2^3 y^3)^2
So,
x^6 - 64y^6 = (x^3 - 8y^3)(x^3 + 8y^3)
we know to to factorize
a^3+ Y^3 = (a + b)(a^2 - ab + b^2 )

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

Substitute

Answer 3

x^6 - 64y^6 =
x^6 - 2^6*y^6 =
x^6 - (2y)^6 =

Let x^6 = a^2 & (2y)^6 = b^2
x^3 = a & (2y)^3 = b

Then
a^2 - b^2 = (a - b)(a + b)
Substituting back in
(x^3 - (2y)^3)(x^3 + (2y)^3)
(x^3 - 8y^3)(x^3 + 8y^3)

Answer 4

x^6 - 64y^6

note that 64 = 2^6
hence
x^6 -(2^6)y^6 = x^6 - (2y)^6

let u=x^3 and v = (2y)^3
then u^2 = (x^3)^2 = x^6
and v^2 = ((2y)^3)^2 = (2y) ^6

we know the identity: u^2 -v^2 = (u-v)(u+v)
Hence,
x^6 - 64y^6 = {x^3-(2y)^3}{x^3+(2y)^3}

and also :
a^3-b^3 = (a-b)(a^2 +ab +b^2)
and
a^3+b^3 = (a+b)(a^2-ab+b^2)

plugging these to get

x^6 - 64y^6 = (x-2y)(x+2y){x^2 +2xy+4y^2)(x^2-2xy +4y^2)

Answer 5

(x^3+8y^3)(x^3-8y^3)

Just foil that and you'll get x^6 - 64y^6