Solve (x^6-2)^2=1

$$( x ^ { 6 } - 2 ) ^ { 2 } = 1$$

Solve for x

$$x=-1$$

$$x=1$$

$$x=\sqrt[6]{3}\approx 1.200936955$$

$$x=-\sqrt[6]{3}\approx -1.200936955$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{6}-2\right)^{2}.

$$\left(x^{6}\right)^{2}-4x^{6}+4=1$$

To raise a power to another power, multiply the exponents. Multiply 6 and 2 to get 12.

$$x^{12}-4x^{6}+4=1$$

Subtract 1 from both sides.

$$x^{12}-4x^{6}+4-1=0$$

Subtract 1 from 4 to get 3.

$$x^{12}-4x^{6}+3=0$$

Substitute t for x^{6}.

$$t^{2}-4t+3=0$$

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, -4 for b, and 3 for c in the quadratic formula.

$$t=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 3}}{2}$$

Do the calculations.

$$t=\frac{4±2}{2}$$

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

$$t=3 t=1$$

Since x=t^{6}, the solutions are obtained by evaluating x=±\sqrt[6]{t} for positive t.

$$x=\sqrt[6]{3} x=-\sqrt[6]{3} x=1 x=-1$$