Factor. \[ 27 u^{3}+w^{3} \] \( \square \)

1. Recognize that \( 27u^3 \) and \( w^3 \) are perfect cubes because: \[ 27u^3 = (3u)^3 \quad \text{and} \quad w^3 = w^3. \] 2. Write the expression as a sum of cubes: \[ 27u^3 + w^3 = (3u)^3 + w^3. \] 3. Use the sum of cubes formula: \[ a^3+b^3 = (a+b)(a^2-ab+b^2). \] Here, \( a = 3u \) and \( b = w \). 4. Substitute into the formula: \[ (3u)^3+w^3 = (3u+w)\left((3u)^2-3u\cdot w+w^2\right). \] 5. Simplify the expression inside the parentheses: \[ (3u)^2 = 9u^2, \] so the factorization becomes: \[ (3u+w)(9u^2-3uw+w^2). \] Thus, the factorization of \( 27u^3+w^3 \) is: \[ (3u+w)(9u^2-3uw+w^2). \]