Solve 8x^6-9x^3+1=0

$$8 x ^ { 6 } - 9 x ^ { 3 } + 1 = 0$$

Solve for x (complex solution)

$$x\in 1,\frac{-1+\sqrt{3}i}{2},\frac{-\sqrt{3}i-1}{2},\frac{-1+\sqrt{3}i}{4},\frac{1}{2},\frac{-\sqrt{3}i-1}{4}$$

Steps Using the Quadratic Formula

Substitute t for x^{3}.

$$8t^{2}-9t+1=0$$

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 8 for a, -9 for b, and 1 for c in the quadratic formula.

$$t=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 8\times 1}}{2\times 8}$$

Do the calculations.

$$t=\frac{9±7}{16}$$

Solve the equation t=\frac{9±7}{16} when ± is plus and when ± is minus.

$$t=1 t=\frac{1}{8}$$

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

$$x=\frac{-1+\sqrt{3}i}{2} x=\frac{-\sqrt{3}i-1}{2} x=1 x=\frac{-1+\sqrt{3}i}{4} x=\frac{-\sqrt{3}i-1}{4} x=\frac{1}{2}$$

Solve for x

$$x=\frac{1}{2}=0.5$$

$$x=1$$

Steps Using the Quadratic Formula

Substitute t for x^{3}.

$$8t^{2}-9t+1=0$$

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 8 for a, -9 for b, and 1 for c in the quadratic formula.

$$t=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 8\times 1}}{2\times 8}$$

Do the calculations.

$$t=\frac{9±7}{16}$$

Solve the equation t=\frac{9±7}{16} when ± is plus and when ± is minus.

$$t=1 t=\frac{1}{8}$$

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

$$x=1 x=\frac{1}{2}$$