How do you factor the expression #81x^6+24x^3y^3#?

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Answer 1 · 2018-04-10T08:24:15

#81x^6 + 24x^3y^3 = 3x^3(3x+2y)(9x^2-6xy+4y^2)#

Given:

#81x^6 + 24x^3y^3#

Note that both factors are divisible by #3x^3#. So we can separate out that factor first to find:

#81x^6 + 24x^3y^3 = 3x^3(27x^3+8y^3)#

Next note that both #27x^3 = (3x)^3# and #8y^3 = (2y)^3# are perfect cubes. So we can use the sum of cubes identity:

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

with #A=3x# and #B=2y# to find:

#27x^3+8y^3 = (3x)^3+(2y)^3#

#color(white)(27x^3+8y^3) = (3x+2y)((3x)^2-(3x)(2y)+(2y)^2)#

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

Putting it all together:

#81x^6 + 24x^3y^3 = 3x^3(3x+2y)(9x^2-6xy+4y^2)#