factor completely x^3 - x^3y^4?

Question

I think the answer is y^4

But i could be missing something?

I think the x^3 can be factored out?

Answer 1

x ³ ( 1 - y^4 )
x ³ ( 1 - y ² ) ( 1 + y ² )
x ³ ( 1 - y ) ( 1 + y ) ( 1 + y ² )

Answer 2

When you factor out, you don't ignore things.

x^3 - (x^3)(y^4) = (x^3)(1 - y^4).
You notice 1 - y^4 is a difference of perfect squares, so that can be further factored.
(x^3)(1 - y^4) = (x^3)(1 - y^2)(1 + y^2).
(1 - y^2) is another difference of perfect squares.
(x^3)(1 - y^2)(1 + y^2) = (x^3)(1 - y)(1 + y)(1 + y^2).

THAT is x^3 - x^3y^4 factored completely.

Answer 3

approach : a^3 - b^3 = (a-b)(a^2+ab+b^2) decision: x^3 - a million/8 making use of approach (x - 0.5)(x^2+x*a million/2+a million/4) now, each and every x-0.5 =0 x=a million/2 or x^2+x*a million/2+a million/4=0 4x^2+2x+a million=0 (multiplying all factors with the aid of 4) by utilising quadratic approach x = -a million/4+root.3/4i , -a million/4-root.3/4i. consequently; decision set a million/2 of , -a million/4+root.3/4i , -a million/4-root.3/4i

Answer 4

Simplifying
x^3 + -1x^3y^4

Factor out the Greatest Common Factor (GCF), 'x3'.
x^3(1 + -1y^4)

Factor a difference between two squares.
x^3((1 + y^2)(1 + -1y^2))

Factor a difference between two squares.
x^3((1 + y^2)((1 + y)(1 + -1y)))

Final result:
x^3(1 + y^2)(1 + y)(1 + -1y)