Answer
-1, 0, and 1
Work Step by Step
First, factor the polynomial completely:
$\\y=3x(x^2-1)
\\y=3x(x-1)(x+1).$
Equate each unique factor to zero, then solve each equation:
$3x=0 \text{ or } x-1=0 \text{ or } x+1=0
\\x=0 \text{ or } x = 1 \text{ or } x=-1.$
Thus, the zeros are -1, 0, and 1, each with multiplicity one.