How do you factor the expression #x^9-27y^6#?

1 Answer
George C.
Apr 7, 2018

#x^9-27y^6 = (x^3-3y^2)(x^6+3x^3y^2+9y^4)#

Explanation:

Given:

#x^9-27y^6#

Note that both #x^9 = (x^3)^3# and #27y^6 = (3y^2)^3# are perfect cubes.

So we can factor the given expression as a difference of cubes:

#A^3-B^3 = (A-B)(A^2+AB+B^2)#

with #A=x^3# and #B=3y^2# as follows:

#x^9-27y^6 = (x^3)^3-(3y^2)^3#

#color(white)(x^9-27y^6) = (x^3-3y^2)((x^3)^2+(x^3)(3y^2)+(3y^2)^2)#

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

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