Solve {l}{1/x^3-3x^2+2x=[x-2/x(x-1)}{=[x-2/x(x-1)}

$$\left. \begin{array}{l}{ \frac { 1 } { x ^ { 3 } - 3 x ^ { 2 } + 2 x } = [ \frac { x - 2 } { x ( x - 1 ) } }\\{ = [ \frac { x - 2 } { x ( x - 1 ) } }\end{array} \right.$$

Solve for x

$$x=3$$

Steps Using Factoring

Variable x cannot be equal to any of the values 0,1,2 since division by zero is not defined. Multiply both sides of the equation by x\left(x-2\right)\left(x-1\right), the least common multiple of x^{3}-3x^{2}+2x,x\left(x-1\right).

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

Multiply x-2 and x-2 to get \left(x-2\right)^{2}.

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

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

$$1=x^{2}-4x+4$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-4x+4=1$$

Subtract 1 from both sides.

$$x^{2}-4x+4-1=0$$

Subtract 1 from 4 to get 3.

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

To solve the equation, factor x^{2}-4x+3 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-4 ab=3$$

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

$$a=-3 b=-1$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-3\right)\left(x-1\right)$$

To find equation solutions, solve x-3=0 and x-1=0.

$$x=3 x=1$$

Variable x cannot be equal to 1.

$$x=3$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values 0,1,2 since division by zero is not defined. Multiply both sides of the equation by x\left(x-2\right)\left(x-1\right), the least common multiple of x^{3}-3x^{2}+2x,x\left(x-1\right).

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

Multiply x-2 and x-2 to get \left(x-2\right)^{2}.

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

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

$$1=x^{2}-4x+4$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-4x+4=1$$

Subtract 1 from both sides.

$$x^{2}-4x+4-1=0$$

Subtract 1 from 4 to get 3.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\times 3=3$$

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

$$a=-3 b=-1$$

Rewrite x^{2}-4x+3 as \left(x^{2}-3x\right)+\left(-x+3\right).

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

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

$$x\left(x-3\right)-\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x-1\right)$$

To find equation solutions, solve x-3=0 and x-1=0.

$$x=3 x=1$$

Variable x cannot be equal to 1.

$$x=3$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,1,2 since division by zero is not defined. Multiply both sides of the equation by x\left(x-2\right)\left(x-1\right), the least common multiple of x^{3}-3x^{2}+2x,x\left(x-1\right).

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

Multiply x-2 and x-2 to get \left(x-2\right)^{2}.

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

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

$$1=x^{2}-4x+4$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-4x+4=1$$

Subtract 1 from both sides.

$$x^{2}-4x+4-1=0$$

Subtract 1 from 4 to get 3.

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

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

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

Square -4.

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

Multiply -4 times 3.

$$x=\frac{-\left(-4\right)±\sqrt{16-12}}{2}$$

Add 16 to -12.

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

Take the square root of 4.

$$x=\frac{-\left(-4\right)±2}{2}$$

The opposite of -4 is 4.

$$x=\frac{4±2}{2}$$

Now solve the equation x=\frac{4±2}{2} when ± is plus. Add 4 to 2.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

Now solve the equation x=\frac{4±2}{2} when ± is minus. Subtract 2 from 4.

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

Divide 2 by 2.

$$x=1$$

The equation is now solved.

$$x=3 x=1$$

Variable x cannot be equal to 1.

$$x=3$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,1,2 since division by zero is not defined. Multiply both sides of the equation by x\left(x-2\right)\left(x-1\right), the least common multiple of x^{3}-3x^{2}+2x,x\left(x-1\right).

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

Multiply x-2 and x-2 to get \left(x-2\right)^{2}.

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

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

$$1=x^{2}-4x+4$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-4x+4=1$$

Factor x^{2}-4x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=1 x-2=-1$$

Add 2 to both sides of the equation.

$$x=3 x=1$$

Variable x cannot be equal to 1.

$$x=3$$