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

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Answer 1 · 2016-01-15T03:45:02

#27x^3-8y^3=(3x)^3-(2y)^3=color(red)((3x-2y))color(red)((9x^2+6xy+4y^2))#

#27x^3-8y^3# fits the form of the difference of cubes, #a^3-b^3#, where #a=3x# and #b=2y#.

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

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

Substitute the known values for #a# and #b#.

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

Simplify.

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