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

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

Solve for x

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

$$x=1$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-1\right)^{3}.

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

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

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

Subtract x^{3} from both sides.

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

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

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

Add x^{2} to both sides.

$$-3x^{2}+3x-1+x^{2}=0$$

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

$$-2x^{2}+3x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=3 ab=-2\left(-1\right)=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=2 b=1$$

Rewrite -2x^{2}+3x-1 as \left(-2x^{2}+2x\right)+\left(x-1\right).

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

Factor out 2x in the first and -1 in the second group.

$$2x\left(-x+1\right)-\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(2x-1\right)$$

To find equation solutions, solve -x+1=0 and 2x-1=0.

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

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-1\right)^{3}.

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

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

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

Subtract x^{3} from both sides.

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

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

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

Add x^{2} to both sides.

$$-3x^{2}+3x-1+x^{2}=0$$

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

$$-2x^{2}+3x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 3 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-2\right)\left(-1\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-3±\sqrt{9+8\left(-1\right)}}{2\left(-2\right)}$$

Multiply 8 times -1.

$$x=\frac{-3±\sqrt{9-8}}{2\left(-2\right)}$$

Add 9 to -8.

$$x=\frac{-3±\sqrt{1}}{2\left(-2\right)}$$

Take the square root of 1.

$$x=\frac{-3±1}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-3±1}{-4}$$

Now solve the equation x=\frac{-3±1}{-4} when ± is plus. Add -3 to 1.

$$x=-\frac{2}{-4}$$

Reduce the fraction \frac{-2}{-4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-3±1}{-4} when ± is minus. Subtract 1 from -3.

$$x=-\frac{4}{-4}$$

Divide -4 by -4.

$$x=1$$

The equation is now solved.

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

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-1\right)^{3}.

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

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

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

Subtract x^{3} from both sides.

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

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

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

Add x^{2} to both sides.

$$-3x^{2}+3x-1+x^{2}=0$$

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

$$-2x^{2}+3x-1=0$$

Add 1 to both sides. Anything plus zero gives itself.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 3 by -2.

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

Divide 1 by -2.

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

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=-\frac{1}{2}+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{1}{2}+\frac{9}{16}$$

Add -\frac{1}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{1}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{1}{4} x-\frac{3}{4}=-\frac{1}{4}$$

Add \frac{3}{4} to both sides of the equation.

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