Solve x^3-left(x-2right)^3=1

$${ x }^{ 3 } - { \left(x-2 \right) }^{ 3 } =1$$

Solve for x (complex solution)

$$x=\frac{\sqrt{6}i}{6}+1\approx 1+0.40824829i$$

$$x=-\frac{\sqrt{6}i}{6}+1\approx 1-0.40824829i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-2\right)^{3}.

$$x^{3}-\left(x^{3}-6x^{2}+12x-8\right)=1$$

To find the opposite of x^{3}-6x^{2}+12x-8, find the opposite of each term.

$$x^{3}-x^{3}+6x^{2}-12x+8=1$$

Combine x^{3} and -x^{3} to get 0.

$$6x^{2}-12x+8=1$$

Subtract 1 from both sides.

$$6x^{2}-12x+8-1=0$$

Subtract 1 from 8 to get 7.

$$6x^{2}-12x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -12 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 6\times 7}}{2\times 6}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 6\times 7}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-12\right)±\sqrt{144-24\times 7}}{2\times 6}$$

Multiply -24 times 7.

$$x=\frac{-\left(-12\right)±\sqrt{144-168}}{2\times 6}$$

Add 144 to -168.

$$x=\frac{-\left(-12\right)±\sqrt{-24}}{2\times 6}$$

Take the square root of -24.

$$x=\frac{-\left(-12\right)±2\sqrt{6}i}{2\times 6}$$

The opposite of -12 is 12.

$$x=\frac{12±2\sqrt{6}i}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{12±2\sqrt{6}i}{12}$$

Now solve the equation x=\frac{12±2\sqrt{6}i}{12} when ± is plus. Add 12 to 2i\sqrt{6}.

$$x=\frac{12+2\sqrt{6}i}{12}$$

Divide 12+2i\sqrt{6} by 12.

$$x=\frac{\sqrt{6}i}{6}+1$$

Now solve the equation x=\frac{12±2\sqrt{6}i}{12} when ± is minus. Subtract 2i\sqrt{6} from 12.

$$x=\frac{-2\sqrt{6}i+12}{12}$$

Divide 12-2i\sqrt{6} by 12.

$$x=-\frac{\sqrt{6}i}{6}+1$$

The equation is now solved.

$$x=\frac{\sqrt{6}i}{6}+1 x=-\frac{\sqrt{6}i}{6}+1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-2\right)^{3}.

$$x^{3}-\left(x^{3}-6x^{2}+12x-8\right)=1$$

To find the opposite of x^{3}-6x^{2}+12x-8, find the opposite of each term.

$$x^{3}-x^{3}+6x^{2}-12x+8=1$$

Combine x^{3} and -x^{3} to get 0.

$$6x^{2}-12x+8=1$$

Subtract 8 from both sides.

$$6x^{2}-12x=1-8$$

Subtract 8 from 1 to get -7.

$$6x^{2}-12x=-7$$

Divide both sides by 6.

$$\frac{6x^{2}-12x}{6}=-\frac{7}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\left(-\frac{12}{6}\right)x=-\frac{7}{6}$$

Divide -12 by 6.

$$x^{2}-2x=-\frac{7}{6}$$

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-2x+1=-\frac{7}{6}+1$$

Add -\frac{7}{6} to 1.

$$x^{2}-2x+1=-\frac{1}{6}$$

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{-\frac{1}{6}}$$

Simplify.

$$x-1=\frac{\sqrt{6}i}{6} x-1=-\frac{\sqrt{6}i}{6}$$

Add 1 to both sides of the equation.

$$x=\frac{\sqrt{6}i}{6}+1 x=-\frac{\sqrt{6}i}{6}+1$$