6) \( 27 x^{3}-1=0 \)

To solve the equation \( 27x^3 - 1 = 0 \), we can factor it using the difference of cubes formula, \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \). Here, we have \( a = 3x \) and \( b = 1 \). Thus, we rewrite the equation as: \[ (3x)^3 - 1^3 = 0 \] Factoring it gives us: \[ (3x - 1)((3x)^2 + 3x \cdot 1 + 1^2) = 0 \] This breaks down to: \[ (3x - 1)(9x^2 + 3x + 1) = 0 \] Setting each factor to zero, we first solve \( 3x - 1 = 0 \), leading to \( x = \frac{1}{3} \). The other factor, \( 9x^2 + 3x + 1 = 0 \), can be solved using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substituting \( a = 9, b = 3, c = 1 \): \[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 9 \cdot 1}}{2 \cdot 9} = \frac{-3 \pm \sqrt{9 - 36}}{18} = \frac{-3 \pm \sqrt{-27}}{18} = \frac{-3 \pm 3i\sqrt{3}}{18} = \frac{-1 \pm i\sqrt{3}}{6} \] So, the complete solutions for the equation \( 27x^3 - 1 = 0 \) are: \[ x = \frac{1}{3}, \quad x = \frac{-1 + i\sqrt{3}}{6}, \quad x = \frac{-1 - i\sqrt{3}}{6} \] Now, that's a sweet mix of real and complex numbers!