Solve =x^6+12x^3=40

$$= x ^ { 6 } + 12 x ^ { 3 } = 40$$

Solve for x (complex solution)

$$x\in \sqrt[3]{2\left(\sqrt{19}+3\right)}e^{\frac{\pi i}{3}},\sqrt[3]{2\left(\sqrt{19}+3\right)}e^{\frac{5\pi i}{3}},-\sqrt[3]{2\left(\sqrt{19}+3\right)},\sqrt[3]{2\left(\sqrt{19}-3\right)}e^{\frac{4\pi i}{3}},\sqrt[3]{2\left(\sqrt{19}-3\right)},\sqrt[3]{2\left(\sqrt{19}-3\right)}e^{\frac{2\pi i}{3}}$$

Steps Using the Quadratic Formula

Subtract 40 from both sides.

$$x^{6}+12x^{3}-40=0$$

Substitute t for x^{3}.

$$t^{2}+12t-40=0$$

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, 12 for b, and -40 for c in the quadratic formula.

$$t=\frac{-12±\sqrt{12^{2}-4\times 1\left(-40\right)}}{2}$$

Do the calculations.

$$t=\frac{-12±4\sqrt{19}}{2}$$

Solve the equation t=\frac{-12±4\sqrt{19}}{2} when ± is plus and when ± is minus.

$$t=2\sqrt{19}-6 t=-2\sqrt{19}-6$$

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

$$x=-\sqrt[3]{2\sqrt{19}-6}e^{\frac{\pi i}{3}} x=\sqrt[3]{2\sqrt{19}-6}ie^{\frac{\pi i}{6}} x=\sqrt[3]{2\sqrt{19}-6} x=-\sqrt[3]{2\sqrt{19}+6}ie^{\frac{\pi i}{6}} x=-\sqrt[3]{2\sqrt{19}+6} x=\sqrt[3]{2\sqrt{19}+6}e^{\frac{\pi i}{3}}$$

Solve for x

$$x=\sqrt[3]{2\sqrt{19}-6}\approx 1.395529599$$

$$x=\sqrt[3]{-2\sqrt{19}-6}\approx -2.450648052$$

Steps Using the Quadratic Formula

Subtract 40 from both sides.

$$x^{6}+12x^{3}-40=0$$

Substitute t for x^{3}.

$$t^{2}+12t-40=0$$

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, 12 for b, and -40 for c in the quadratic formula.

$$t=\frac{-12±\sqrt{12^{2}-4\times 1\left(-40\right)}}{2}$$

Do the calculations.

$$t=\frac{-12±4\sqrt{19}}{2}$$

Solve the equation t=\frac{-12±4\sqrt{19}}{2} when ± is plus and when ± is minus.

$$t=2\sqrt{19}-6 t=-2\sqrt{19}-6$$

Since x=t^{3}, the solutions are obtained by evaluating x=\sqrt[3]{t} for each t.

$$x=\sqrt[3]{2\sqrt{19}-6} x=\sqrt[3]{-2\sqrt{19}-6}$$