## Intermediate Algebra for College Students (7th Edition)

$(3y+1)(9y^2-3y+1)$
The given binomial can be written as: $(3y)^3+1^3$ This binomial is a sum of two cubes. RECALL: $a^3+b^3=(a+b)(a^2-ab+b^2)$ Factor the difference of two cubes using the formula above with $a=3y$ and $b=1$ to obtain: $=(3y+1)[(3y)^2-3y(1)+1^2) \\=(3y+1)(9y^2-3y+1)$