Factor completely. \[ 9 y^{2}-6 y+1 \]

To factor the expression \(9y^2 - 6y + 1\), we can use the technique of recognizing it as a perfect square trinomial. Notice that \(9y^2\) is \((3y)^2\), and \(1\) is \(1^2\). The middle term’s coefficient, \(-6y\), can be expressed as \(-2 \cdot 3y \cdot 1\). Thus, we can rewrite the expression as: \[ (3y - 1)^2 \] Now, let’s do a quick check to ensure this is correct by expanding \((3y - 1)(3y - 1)\): \[ (3y - 1)(3y - 1) = 3y \cdot 3y - 3y \cdot 1 - 1 \cdot 3y + 1 \cdot 1 = 9y^2 - 3y - 3y + 1 = 9y^2 - 6y + 1 \] Thus, the complete factorization is: \[ (3y - 1)^2 \]