How do you factor the expression #27x^9 + 8y^6 #?

1 Answer
Shwetank Mauria
May 28, 2016

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

Explanation:

#27x^9+8y^6# is a binomial each of whose monomial is a cube. Hence we can use the identity

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

As such #27x^9+8y^6=(3x^3)^3+(2y^2)^3#

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

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

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