First Principles AI

Solution

\begin{aligned} & \boxed{\text{We know the sum of cubes formula:}} \\ \\ & a^3 + b^3 = (a + b)(a^2 - ab + b^2) \\ \\ & \text{So, } \\ & z^3 + 27 = z^3 + (3)^3 \\ & a = z, \quad b = 3 \\ \\ & \boxed{\text{Applying the formula:}} \\ \\ & z^3 + 27 = (z + 3)(z^2 - 3z + 9) \\ \\ & \boxed{\text{Now factor } z^2 - 3z + 9:} \\ & a = 1, \; b = -3, \; c = 9 \\ & \Delta = b^2 - 4ac = (-3)^2 - 4(1)(9) = 9 - 36 = -27 \\ & \sqrt{\Delta} = \sqrt{-27} = 3i\sqrt{3} \\ \\ & \boxed{\text{Using the quadratic formula:}} \\ & z = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-(-3) \pm 3i\sqrt{3}}{2} = \frac{3 \pm 3i\sqrt{3}}{2} \\ & z_1 = \frac{3 + 3i\sqrt{3}}{2}, \quad z_2 = \frac{3 - 3i\sqrt{3}}{2} \\ \\ & \boxed{\text{Therefore, the factors are:}} \\ & z^2 - 3z + 9 = (z - z_1)(z - z_2) \\ & = \left(z - \frac{3 + 3i\sqrt{3}}{2}\right) \left(z - \frac{3 - 3i\sqrt{3}}{2}\right) \\ \\ & \boxed{\text{Therefore,}} \\ & z^3 + 27 = (z + 3)\left(z - \frac{3 + 3i\sqrt{3}}{2}\right)\left(z - \frac{3 - 3i\sqrt{3}}{2}\right) \\ \end{aligned}