Solve x^3/x^3-1-1/2x-2=1

$$\frac { x ^ { 3 } } { x ^ { 3 } - 1 } - \frac { 1 } { 2 x - 2 } = 1$$

Solve for x

$$x=\frac{\sqrt{5}-1}{2}\approx 0.618033989$$

$$x=\frac{-\sqrt{5}-1}{2}\approx -1.618033989$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-1\right)\left(x^{2}+x+1\right), the least common multiple of x^{3}-1,2x-2.

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

To find the opposite of x^{2}+x+1, find the opposite of each term.

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

Use the distributive property to multiply 2 by x-1.

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

Use the distributive property to multiply 2x-2 by x^{2}+x+1 and combine like terms.

$$2x^{3}-x^{2}-x-1=2x^{3}-2$$

Subtract 2x^{3} from both sides.

$$2x^{3}-x^{2}-x-1-2x^{3}=-2$$

Combine 2x^{3} and -2x^{3} to get 0.

$$-x^{2}-x-1=-2$$

Add 2 to both sides.

$$-x^{2}-x-1+2=0$$

Add -1 and 2 to get 1.

$$-x^{2}-x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -1 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Multiply -4 times -1.

$$x=\frac{-\left(-1\right)±\sqrt{1+4}}{2\left(-1\right)}$$

Add 1 to 4.

$$x=\frac{-\left(-1\right)±\sqrt{5}}{2\left(-1\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{5}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{1±\sqrt{5}}{-2}$$

Now solve the equation x=\frac{1±\sqrt{5}}{-2} when ± is plus. Add 1 to \sqrt{5}.

$$x=\frac{\sqrt{5}+1}{-2}$$

Divide 1+\sqrt{5} by -2.

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

Now solve the equation x=\frac{1±\sqrt{5}}{-2} when ± is minus. Subtract \sqrt{5} from 1.

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

Divide 1-\sqrt{5} by -2.

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

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-1\right)\left(x^{2}+x+1\right), the least common multiple of x^{3}-1,2x-2.

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

To find the opposite of x^{2}+x+1, find the opposite of each term.

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

Use the distributive property to multiply 2 by x-1.

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

Use the distributive property to multiply 2x-2 by x^{2}+x+1 and combine like terms.

$$2x^{3}-x^{2}-x-1=2x^{3}-2$$

Subtract 2x^{3} from both sides.

$$2x^{3}-x^{2}-x-1-2x^{3}=-2$$

Combine 2x^{3} and -2x^{3} to get 0.

$$-x^{2}-x-1=-2$$

Add 1 to both sides.

$$-x^{2}-x=-2+1$$

Add -2 and 1 to get -1.

$$-x^{2}-x=-1$$

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide -1 by -1.

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

Divide -1 by -1.

$$x^{2}+x=1$$

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

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=1+\frac{1}{4}$$

Add 1 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{5}{4}$$

Factor x^{2}+x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{5}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{5}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{5}}{2} x+\frac{1}{2}=-\frac{\sqrt{5}}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

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