Solve x^6-x^3+4=2

$$x ^ { 6 } - x ^ { 3 } + 4 = 2$$

Solve for x (complex solution)

$$x\in \sqrt[6]{2}e^{\frac{-\arctan(\sqrt{7})i+2\pi i}{3}},\sqrt[6]{2}e^{-\frac{\arctan(\sqrt{7})i}{3}},\sqrt[6]{2}e^{\frac{-\arctan(\sqrt{7})i+4\pi i}{3}},\sqrt[6]{2}e^{\frac{\arctan(\sqrt{7})i+4\pi i}{3}},\sqrt[6]{2}e^{\frac{\arctan(\sqrt{7})i}{3}},\sqrt[6]{2}e^{\frac{\arctan(\sqrt{7})i+2\pi i}{3}}$$

Steps Using the Quadratic Formula

Subtract 2 from both sides.

$$x^{6}-x^{3}+4-2=0$$

Subtract 2 from 4 to get 2.

$$x^{6}-x^{3}+2=0$$

Substitute t for x^{3}.

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

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 2 for c in the quadratic formula.

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

Do the calculations.

$$t=\frac{1±\sqrt{-7}}{2}$$

Solve the equation t=\frac{1±\sqrt{-7}}{2} when ± is plus and when ± is minus.

$$t=\frac{1+\sqrt{7}i}{2} t=\frac{-\sqrt{7}i+1}{2}$$

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

$$x=\sqrt[6]{2}e^{\frac{\arctan(\sqrt{7})i+4\pi i}{3}} x=\sqrt[6]{2}e^{\frac{\arctan(\sqrt{7})i+2\pi i}{3}} x=\sqrt[6]{2}e^{\frac{\arctan(\sqrt{7})i}{3}} x=\sqrt[6]{2}e^{-\frac{\arctan(\sqrt{7})i}{3}} x=\sqrt[6]{2}e^{\frac{-\arctan(\sqrt{7})i+4\pi i}{3}} x=\sqrt[6]{2}e^{\frac{-\arctan(\sqrt{7})i+2\pi i}{3}}$$