Question

how do you factor 27y^3+8

Answer 1

thank you

Answer 2

\[\large 27y^3+8=(3y)^3+2^3=(3y+2)\cdot[(3y)^2-(3y)2+2^2] \]

Answer 3

it is the sum of two cubes, so it can be factored following
\[\large a^3+b^3=(a+b)(a^2-ab+b^2) \]

Answer 4

Not a nice factorization.