How do you factor x^3 - 8y^3x3−8y3?

3 Answers
Nityananda
Mar 29, 2018

(x-2y)(x^2+2xy+4y^2)(x−2y)(x2+2xy+4y2)

Explanation:

x^3-8y^3x3−8y3
rArr (x)^3-(2y)^3⇒(x)3−(2y)3
rArr(x-2y)^3+3.x.2y(x-2y)⇒(x−2y)3+3.x.2y(x−2y)
rArr(x-2y)[(x-2y)^2+6xy]⇒(x−2y)[(x−2y)2+6xy]
rArr(x-2y)[x^2-4xy+4y^2+6xy]⇒(x−2y)[x2−4xy+4y2+6xy]
rArr(x-2y)(x^2+2xy+4y^2)⇒(x−2y)(x2+2xy+4y2)

Jim G.
Mar 29, 2018

(x-2y)(x^2+2xy+4y^2)(x−2y)(x2+2xy+4y2)

Explanation:

x^3-8y^3larrcolor(blue)"is a difference of cubes"x3−8y3←is a difference of cubes

•color(white)(x)a^3-b^3=(a-b)(a^2+ab+b^2)∙xa3−b3=(a−b)(a2+ab+b2)

"here "a=x" and "b=2yto(2y)^3=8y^3here a=x and b=2y→(2y)3=8y3

rArrx^3-8y^3=(x-2y)(x^2+2xy+4y^2)⇒x3−8y3=(x−2y)(x2+2xy+4y2)

Meave60
Mar 29, 2018

x^3-8y^3=color(blue)((x-2y)(x^2+2xy+4y^2)x3−8y3=(x−2y)(x2+2xy+4y2)

Explanation:

Factor:

x^3-8y^3x3−8y3

Apply the difference of cubes:

(a^2-b^3)=(a-b)(a^2+ab+b^2)(a2−b3)=(a−b)(a2+ab+b2),

where:

a=xa=x, and b=2yb=2y

Plug in the known values.

x^3-(2y)^3=x3−(2y)3=

(x-2y)(x^2+(x)(2y)+(2y)^2)(x−2y)(x2+(x)(2y)+(2y)2)

Apply multiplicative distributive property: (ab)^m=a^mb^m(ab)m=ambm

(x-2y)(x^2+(x)(2y)+(2^2y^2))(x−2y)(x2+(x)(2y)+(22y2))

Simplify.

(x-2y)(x^2+2xy+4y^2)(x−2y)(x2+2xy+4y2)

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