Khi phân tích đa thức \(27x^3-\dfrac{1}{8}\) thành nhân tử, ta có kết quả là
\(\left(3x-\dfrac{1}{2}\right)\left(9x^2+\dfrac{3}{2}x+\dfrac{1}{4}\right)\).\(\left(3x-\dfrac{1}{2}\right)\left(9x^2-\dfrac{3}{2}x+\dfrac{1}{4}\right)\).\(\left(3x+\dfrac{1}{2}\right)\left(9x^2-\dfrac{3}{2}x+\dfrac{1}{4}\right)\).\(\left(3x-\dfrac{1}{2}\right)\left(3x+\dfrac{1}{2}\right)\).Hướng dẫn giải:\(27x^3-\dfrac{1}{8}=\left(3x\right)^3-\left(\dfrac{1}{2}\right)^3=\left(3x-\dfrac{1}{2}\right)\left(9x^2+\dfrac{3}{2}x+\dfrac{1}{4}\right)\)