Solve x^3+x^6=1

$$x ^ { 3 } + x ^ { 6 } = 1$$

Solve for x (complex solution)

$$x\in \frac{2^{\frac{2}{3}}\sqrt[3]{\sqrt{5}+1}e^{\frac{\pi i}{3}}}{2},\frac{2^{\frac{2}{3}}\sqrt[3]{\sqrt{5}+1}e^{\frac{5\pi i}{3}}}{2},-\frac{2^{\frac{2}{3}}\sqrt[3]{\sqrt{5}+1}}{2},\frac{2^{\frac{2}{3}}\sqrt[3]{\sqrt{5}-1}e^{\frac{4\pi i}{3}}}{2},\frac{2^{\frac{2}{3}}\sqrt[3]{\sqrt{5}-1}}{2},\frac{2^{\frac{2}{3}}\sqrt[3]{\sqrt{5}-1}e^{\frac{2\pi i}{3}}}{2}$$

Steps Using the Quadratic Formula

Subtract 1 from both sides.

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

Substitute t for x^{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.

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

Do the calculations.

$$t=\frac{-1±\sqrt{5}}{2}$$

Solve the equation t=\frac{-1±\sqrt{5}}{2} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{5}-1}{2} t=\frac{-\sqrt{5}-1}{2}$$

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

$$x=-\sqrt[3]{\frac{\sqrt{5}-1}{2}}e^{\frac{\pi i}{3}} x=\sqrt[3]{\frac{\sqrt{5}-1}{2}}ie^{\frac{\pi i}{6}} x=\sqrt[3]{\frac{\sqrt{5}-1}{2}} x=-\sqrt[3]{\frac{\sqrt{5}+1}{2}}ie^{\frac{\pi i}{6}} x=-\sqrt[3]{\frac{\sqrt{5}+1}{2}} x=\sqrt[3]{\frac{\sqrt{5}+1}{2}}e^{\frac{\pi i}{3}}$$

Solve for x

$$x=\frac{2^{\frac{2}{3}}\sqrt[3]{\sqrt{5}-1}}{2}\approx 0.851799642$$

$$x = \frac{2 ^ {\frac{2}{3}} \sqrt[3]{-\sqrt{5} - 1}}{2} \approx 1.173984997$$

Steps Using the Quadratic Formula

Subtract 1 from both sides.

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

Substitute t for x^{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.

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

Do the calculations.

$$t=\frac{-1±\sqrt{5}}{2}$$

Solve the equation t=\frac{-1±\sqrt{5}}{2} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{5}-1}{2} t=\frac{-\sqrt{5}-1}{2}$$

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

$$x=\sqrt[3]{\frac{\sqrt{5}-1}{2}} x=\sqrt[3]{\frac{-\sqrt{5}-1}{2}}$$