1. The problem asks to factor the polynomial $1 - 8c^3$. 2. Recognize this as a difference of cubes: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. 3. Here, $1 = 1^3$ and $8c^3 = (2c)^3$, so $a = 1$ and $b = 2c$. 4. Apply the formula: $$1 - 8c^3 = (1 - 2c)(1^2 + 1 \cdot 2c + (2c)^2) = (1 - 2c)(1 + 2c + 4c^2)$$ 5. Rearranging the first factor to match the choices: $(1 - 2c) = -(2c - 1)$, so $$1 - 8c^3 = -(2c - 1)(1 + 2c + 4c^2)$$ 6. Since factoring out a negative is optional, the factorization is: $$(2c - 1)(4c^2 + 2c + 1)$$ --- 7. Next, factor $125k^3 + 27$. 8. Recognize this as a sum of cubes: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$. 9. Here, $125k^3 = (5k)^3$ and $27 = 3^3$, so $a = 5k$ and $b = 3$. 10. Apply the formula: $$125k^3 + 27 = (5k + 3)((5k)^2 - 5k \cdot 3 + 3^2) = (5k + 3)(25k^2 - 15k + 9)$$ Final answers: - $1 - 8c^3 = (2c - 1)(4c^2 + 2c + 1)$ - $125k^3 + 27 = (5k + 3)(25k^2 - 15k + 9)$