\begin{tabular}{l} Factor each completely. \\ \begin{tabular}{ll}\( x^{3}+1 \) & 2) \( 125-8 x^{3} \) \\ 3) \( 8 x^{3}-27 \) & 4) \( u^{3}-64 \) \\ 5) \( u^{3}-1 \) & \\ \hline\end{tabular} \\ \hline\end{tabular}

Certainly! Let's factor each expression completely using the formulas for sum and difference of cubes. ### 1) \( x^{3} + 1 \) This is a **sum of cubes**, which can be factored using the formula: \[ a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2}) \] Here, \( a = x \) and \( b = 1 \). Applying the formula: \[ x^{3} + 1 = (x + 1)(x^{2} - x \cdot 1 + 1^{2}) = (x + 1)(x^{2} - x + 1) \] ### 2) \( 125 - 8x^{3} \) This is a **difference of cubes**, recognizable as \( 5^{3} - (2x)^{3} \). Using the formula: \[ a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2}) \] Let \( a = 5 \) and \( b = 2x \). Thus: \[ 125 - 8x^{3} = (5 - 2x)(5^{2} + 5 \cdot 2x + (2x)^{2}) = (5 - 2x)(25 + 10x + 4x^{2}) \] ### 3) \( 8x^{3} - 27 \) Another **difference of cubes**, where \( (2x)^{3} - 3^{3} \). Applying the difference of cubes formula: \[ 8x^{3} - 27 = (2x - 3)((2x)^{2} + 2x \cdot 3 + 3^{2}) = (2x - 3)(4x^{2} + 6x + 9) \] ### 4) \( u^{3} - 64 \) This is a **difference of cubes** since \( u^{3} - 4^{3} \). Using the formula: \[ u^{3} - 64 = (u - 4)(u^{2} + u \cdot 4 + 4^{2}) = (u - 4)(u^{2} + 4u + 16) \] ### 5) \( u^{3} - 1 \) This is a **difference of cubes** with \( u^{3} - 1^{3} \). Applying the formula: \[ u^{3} - 1 = (u - 1)(u^{2} + u \cdot 1 + 1^{2}) = (u - 1)(u^{2} + u + 1) \] ### Summary of Factored Forms: 1. \( x^{3} + 1 = (x + 1)(x^{2} - x + 1) \) 2. \( 125 - 8x^{3} = (5 - 2x)(25 + 10x + 4x^{2}) \) 3. \( 8x^{3} - 27 = (2x - 3)(4x^{2} + 6x + 9) \) 4. \( u^{3} - 64 = (u - 4)(u^{2} + 4u + 16) \) 5. \( u^{3} - 1 = (u - 1)(u^{2} + u + 1) \) These factorizations use the standard formulas for factoring sums and differences of cubes.