How do you factor #27+8x^3#?

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Answer 1 · 2016-02-05T00:23:34

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

The sum of two cubes can be factored as:

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

For the problem at hand, #27+8x^3=3^3+(2x)^3# so that #a=3# and #b=2x#. Therefore

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

Answer 2 · 2016-02-05T00:24:39

It is #27+8x^3= (3+2x)*(9-6x+4x^2)#

Rewrite this as follows

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