Answer
$(y-1)(y^2+y+1)$
Work Step by Step
The expressions $
y^3
$ and $
1
$ are both perfect cubes (the cube root is exact). Hence, $
y^3-1
$ is a $\text{
difference
}$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(y)^3-(1)^3
\\\\=
(y-1)[(y)^2+y(1)+(1)^2]
\\\\=
(y-1)(y^2+y+1)
.\end{array}
