Factor the expression: \( 8y^{3} - 1 \)

To factor the expression \( 8y^{3} - 1 \), we can recognize it as a **difference of cubes**. Here's how to proceed: 1. **Identify the Cubes:** - \( 8y^3 = (2y)^3 \) - \( 1 = 1^3 \) 2. **Apply the Difference of Cubes Formula:** The formula for factoring a difference of cubes is: \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \] Here, \( a = 2y \) and \( b = 1 \). 3. **Substitute into the Formula:** \[ 8y^3 - 1 = (2y)^3 - 1^3 = (2y - 1)\left( (2y)^2 + (2y)(1) + 1^2 \right) \] Simplifying further: \[ (2y - 1)(4y^2 + 2y + 1) \] 4. **Check for Further Factoring:** The quadratic \( 4y^2 + 2y + 1 \) does not factor neatly over the real numbers because its discriminant is negative: \[ \Delta = (2)^2 - 4(4)(1) = 4 - 16 = -12 \] Since the discriminant is negative, the quadratic cannot be factored further using real numbers. **Final Factored Form:** \[ 8y^{3} - 1 = (2y - 1)(4y^2 + 2y + 1) \] **Answer:** After factoring, the expression is (2 y – 1) multiplied by (4y² + 2y + 1). Thus, (2y−1)(4y²+2y+1)