Solve z^6-z^3+1=0

$$z ^ { 6 } - z ^ { 3 } + 1 = 0$$

Solve for z

$$z\in e^{\frac{5\pi i}{9}},e^{\frac{17\pi i}{9}},e^{\frac{11\pi i}{9}},e^{\frac{13\pi i}{9}},e^{\frac{\pi i}{9}},e^{\frac{7\pi i}{9}}$$

Steps Using the Quadratic Formula

Substitute t for z^{3}.

$$t^{2}-t+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -1 for b, and 1 for c in the quadratic formula.

$$t=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$t=\frac{1±\sqrt{-3}}{2}$$

Solve the equation t=\frac{1±\sqrt{-3}}{2} when ± is plus and when ± is minus.

$$t=\frac{1+\sqrt{3}i}{2} t=\frac{-\sqrt{3}i+1}{2}$$

Since z=t^{3}, the solutions are obtained by solving the equation for each t.

$$z=-e^{\frac{4\pi i}{9}} z=ie^{\frac{5\pi i}{18}} z=e^{\frac{\pi i}{9}} z=-ie^{\frac{7\pi i}{18}} z=-e^{\frac{2\pi i}{9}} z=ie^{\frac{\pi i}{18}}$$