How do you factor #x^3 - 8y^3#?

3 Answers
Mar 29, 2018

Answer:

#(x-2y)(x^2+2xy+4y^2)#

Explanation:

#x^3-8y^3#
#rArr (x)^3-(2y)^3#
#rArr(x-2y)^3+3.x.2y(x-2y)#
#rArr(x-2y)[(x-2y)^2+6xy]#
#rArr(x-2y)[x^2-4xy+4y^2+6xy]#
#rArr(x-2y)(x^2+2xy+4y^2)#

Mar 29, 2018

Answer:

#(x-2y)(x^2+2xy+4y^2)#

Explanation:

#x^3-8y^3larrcolor(blue)"is a difference of cubes"#

#•color(white)(x)a^3-b^3=(a-b)(a^2+ab+b^2)#

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

#rArrx^3-8y^3=(x-2y)(x^2+2xy+4y^2)#

Mar 29, 2018

Answer:

#x^3-8y^3=color(blue)((x-2y)(x^2+2xy+4y^2)#

Explanation:

Factor:

#x^3-8y^3#

Apply the difference of cubes:

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

where:

#a=x#, and #b=2y#

Plug in the known values.

#x^3-(2y)^3=#

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

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

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

Simplify.

#(x-2y)(x^2+2xy+4y^2)#