How do you factor completely #8y^3 + 27#?

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Answer 1 · 2018-03-25T17:28:36

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

Sum of cubes: #a^3+b^3#

Factored sum of cubes: #(a+b)(a^2-ab+b^2)#

In this case, a is #2y# and b is #3#

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

Plug 2y and 3 in for a and b, respectively

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

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