Factor completely. If the polynomial is prime, state so. \( 64 x^{3}-y^{3} \)

To factor the polynomial \( 64x^{3} - y^{3} \) completely, follow these steps: 1. **Recognize the Difference of Cubes:** The given polynomial is a difference of cubes since \( 64x^{3} = (4x)^3 \) and \( y^{3} = y^3 \). The general formula for factoring a difference of cubes is: \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \] 2. **Apply the Formula:** Let \( a = 4x \) and \( b = y \). Applying the formula: \[ (4x)^3 - y^3 = (4x - y)\left((4x)^2 + (4x)(y) + y^2\right) \] Simplify the expression inside the parentheses: \[ = (4x - y)(16x^2 + 4xy + y^2) \] 3. **Check for Further Factorization:** The quadratic \( 16x^2 + 4xy + y^2 \) can be checked for further factorization. Treating it as a quadratic in \( x \): \[ 16x^2 + 4xy + y^2 \] Calculate the discriminant: \[ D = (4y)^2 - 4 \cdot 16 \cdot y^2 = 16y^2 - 64y^2 = -48y^2 \] Since the discriminant is negative (\( D < 0 \)), the quadratic does not factor further over the real numbers. 4. **Final Factored Form:** Therefore, the completely factored form of the polynomial is: \[ 64x^{3} - y^{3} = (4x - y)(16x^2 + 4xy + y^2) \] Since \( 16x^2 + 4xy + y^2 \) cannot be factored further using real numbers, the expression is fully factored. **Final Answer:** \[ 64x^{3} - y^{3} = (4x - y)(16x^{2} + 4xy + y^{2}) \]