(3) \( a^{3}-27 \) (6) \( 27 p^{3}-8 q^{3} \) (9) \( -125+p^{3} \) (12)* \( x^{6}-y^{6} \) (Careful!) (15) \( \frac{x^{3}}{8}+\frac{64}{y^{3}} \)

Sure, let's factor each of the given expressions step by step. --- ### (3) \( a^{3} - 27 \) This is a **difference of cubes** since \( 27 = 3^3 \). \[ a^{3} - 3^{3} = (a - 3)(a^{2} + 3a + 9) \] **Factored Form:** \[ a^{3} - 27 = (a - 3)(a^{2} + 3a + 9) \] --- ### (6) \( 27p^{3} - 8q^{3} \) This is also a **difference of cubes** where \( 27p^{3} = (3p)^3 \) and \( 8q^{3} = (2q)^3 \). \[ (3p)^{3} - (2q)^{3} = (3p - 2q)\left[(3p)^2 + 3p \cdot 2q + (2q)^2\right] \] \[ = (3p - 2q)(9p^{2} + 6pq + 4q^{2}) \] **Factored Form:** \[ 27p^{3} - 8q^{3} = (3p - 2q)(9p^{2} + 6pq + 4q^{2}) \] --- ### (9) \( -125 + p^{3} \) Rewrite the expression to identify it as a **difference of cubes**. Note that \( -125 = (-5)^3 \). \[ p^{3} - 5^{3} = (p - 5)\left[p^{2} + 5p + 25\right] \] **Factored Form:** \[ -125 + p^{3} = (p - 5)(p^{2} + 5p + 25) \] --- ### (12)* \( x^{6} - y^{6} \) *(Careful!)* This expression is a **difference of squares**, which can be further factored into a product of **difference and sum of cubes**. 1. **First, apply the difference of squares:** \[ x^{6} - y^{6} = (x^{3})^{2} - (y^{3})^{2} = (x^{3} - y^{3})(x^{3} + y^{3}) \] 2. **Now, factor each part using the difference and sum of cubes formulas:** \[ x^{3} - y^{3} = (x - y)(x^{2} + xy + y^{2}) \] \[ x^{3} + y^{3} = (x + y)(x^{2} - xy + y^{2}) \] 3. **Combine all factors:** \[ x^{6} - y^{6} = (x - y)(x + y)(x^{2} + xy + y^{2})(x^{2} - xy + y^{2}) \] **Factored Form:** \[ x^{6} - y^{6} = (x - y)(x + y)(x^{2} + xy + y^{2})(x^{2} - xy + y^{2}) \] --- ### (15) \( \frac{x^{3}}{8} + \frac{64}{y^{3}} \) First, rewrite the expression to recognize it as a **sum of cubes**: \[ \frac{x^{3}}{8} + \frac{64}{y^{3}} = \left(\frac{x}{2}\right)^{3} + \left(\frac{4}{y}\right)^{3} \] Now, apply the sum of cubes formula: \[ a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2}) \] Here, \( a = \frac{x}{2} \) and \( b = \frac{4}{y} \): \[ \left(\frac{x}{2} + \frac{4}{y}\right)\left(\left(\frac{x}{2}\right)^{2} - \frac{x}{2} \cdot \frac{4}{y} + \left(\frac{4}{y}\right)^{2}\right) \] Simplify inside the parentheses: \[ = \left(\frac{x}{2} + \frac{4}{y}\right)\left(\frac{x^{2}}{4} - \frac{2x}{y} + \frac{16}{y^{2}}\right) \] **Factored Form:** \[ \frac{x^{3}}{8} + \frac{64}{y^{3}} = \left(\frac{x}{2} + \frac{4}{y}\right)\left(\frac{x^{2}}{4} - \frac{2x}{y} + \frac{16}{y^{2}}\right) \] --- Feel free to reach out if you need further clarification on any of these factorizations!