How do you factor #27x^3-8y^3#?

1 Answer
Apr 1, 2018

Answer:

The factored polynomial is #(3x-2y)(9x^2+6xy+4y^2)#.

Explanation:

You can use the difference of cubes factoring pattern:

#color(red)a^3-color(blue)b^3quad=quad(color(red)a-color(blue)b)(color(red)a^2+color(red)acolor(blue)b+color(blue)b^2)#

Right now, our terms are #a^3=27x^3# and #b^3=8y^3#. Take the cube root of each to find that #a=3x# and #b=2y#. Now plug them into the difference of cubes:

#color(white)=27x^3-8y^3#

#=(color(red)(3x))^3-(color(blue)(2y))^3#

#=(color(red)(3x)-color(blue)(2y))((color(red)(3x))^2+(color(red)(3x))(color(blue)(2y))+(color(blue)(2y))^2)#

#=(color(red)(3x)-color(blue)(2y))(color(red)3^2color(red)x^2+color(purple)6color(red)xcolor(blue)y+color(blue)2^2color(blue)y^2)#

#=(color(red)(3x)-color(blue)(2y))(color(red)(9x)^2+color(purple)6color(red)xcolor(blue)y+color(blue)(4y)^2)#

#=(3x-2y)(9x^2+6xy+4y^2)#

This is factored all the way. Hope this helped!