Answer
$8y^3+1=(2y+1)(4y^2-2y+1)$
Work Step by Step
The given expression can be written as:
$=(2y)^3+1^3$
RECALL:
A sum or difference of two cubes can be factored using the following:
(i) $a^3-b^3=(a-b)(a^2+ab+b^2)$
(ii) $a^3+b^3 = (a+b)(a^2-ab+b^2)$
Use formula (2) above with $a=2y$ and $b=1$ to have:
$=(2y+1)[(2y)^2-(2y(1)+1^2]
\\=(2y+1)(4y^2-2y+1)$
