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

Question

6609 views around the world

You can reuse this answer
Creative Commons License

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

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

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

How do you factor completely 8y^3 + 278y^3 + 27?