Solve (x-1)^3=x^2(x-1)-(x+3)(x-2)-19

$$( x - 1 ) ^ { 3 } = x ^ { 2 } ( x - 1 ) - ( x + 3 ) ( x - 2 ) - 19$$

Solve for x

$$x=-2$$

$$x=6$$

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}.

$$x^{3}-3x^{2}+3x-1=x^{2}\left(x-1\right)-\left(x+3\right)\left(x-2\right)-19$$

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

$$x^{3}-3x^{2}+3x-1=x^{3}-x^{2}-\left(x+3\right)\left(x-2\right)-19$$

Use the distributive property to multiply x+3 by x-2 and combine like terms.

$$x^{3}-3x^{2}+3x-1=x^{3}-x^{2}-\left(x^{2}+x-6\right)-19$$

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

$$x^{3}-3x^{2}+3x-1=x^{3}-x^{2}-x^{2}-x+6-19$$

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

$$x^{3}-3x^{2}+3x-1=x^{3}-2x^{2}-x+6-19$$

Subtract 19 from 6 to get -13.

$$x^{3}-3x^{2}+3x-1=x^{3}-2x^{2}-x-13$$

Subtract x^{3} from both sides.

$$x^{3}-3x^{2}+3x-1-x^{3}=-2x^{2}-x-13$$

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

$$-3x^{2}+3x-1=-2x^{2}-x-13$$

Add 2x^{2} to both sides.

$$-3x^{2}+3x-1+2x^{2}=-x-13$$

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

$$-x^{2}+3x-1=-x-13$$

Add x to both sides.

$$-x^{2}+3x-1+x=-13$$

Combine 3x and x to get 4x.

$$-x^{2}+4x-1=-13$$

Add 13 to both sides.

$$-x^{2}+4x-1+13=0$$

Add -1 and 13 to get 12.

$$-x^{2}+4x+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=4 ab=-12=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 4.

$$a=6 b=-2$$

Rewrite -x^{2}+4x+12 as \left(-x^{2}+6x\right)+\left(-2x+12\right).

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

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

$$-x\left(x-6\right)-2\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(-x-2\right)$$

To find equation solutions, solve x-6=0 and -x-2=0.

$$x=6 x=-2$$

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}.

$$x^{3}-3x^{2}+3x-1=x^{2}\left(x-1\right)-\left(x+3\right)\left(x-2\right)-19$$

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

$$x^{3}-3x^{2}+3x-1=x^{3}-x^{2}-\left(x+3\right)\left(x-2\right)-19$$

Use the distributive property to multiply x+3 by x-2 and combine like terms.

$$x^{3}-3x^{2}+3x-1=x^{3}-x^{2}-\left(x^{2}+x-6\right)-19$$

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

$$x^{3}-3x^{2}+3x-1=x^{3}-x^{2}-x^{2}-x+6-19$$

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

$$x^{3}-3x^{2}+3x-1=x^{3}-2x^{2}-x+6-19$$

Subtract 19 from 6 to get -13.

$$x^{3}-3x^{2}+3x-1=x^{3}-2x^{2}-x-13$$

Subtract x^{3} from both sides.

$$x^{3}-3x^{2}+3x-1-x^{3}=-2x^{2}-x-13$$

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

$$-3x^{2}+3x-1=-2x^{2}-x-13$$

Add 2x^{2} to both sides.

$$-3x^{2}+3x-1+2x^{2}=-x-13$$

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

$$-x^{2}+3x-1=-x-13$$

Add x to both sides.

$$-x^{2}+3x-1+x=-13$$

Combine 3x and x to get 4x.

$$-x^{2}+4x-1=-13$$

Add 13 to both sides.

$$-x^{2}+4x-1+13=0$$

Add -1 and 13 to get 12.

$$-x^{2}+4x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 4 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-4±\sqrt{16+4\times 12}}{2\left(-1\right)}$$

Multiply 4 times 12.

$$x=\frac{-4±\sqrt{16+48}}{2\left(-1\right)}$$

Add 16 to 48.

$$x=\frac{-4±\sqrt{64}}{2\left(-1\right)}$$

Take the square root of 64.

$$x=\frac{-4±8}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-4±8}{-2}$$

Now solve the equation x=\frac{-4±8}{-2} when ± is plus. Add -4 to 8.

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

Divide 4 by -2.

$$x=-2$$

Now solve the equation x=\frac{-4±8}{-2} when ± is minus. Subtract 8 from -4.

$$x=-\frac{12}{-2}$$

Divide -12 by -2.

$$x=6$$

The equation is now solved.

$$x=-2 x=6$$

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}.

$$x^{3}-3x^{2}+3x-1=x^{2}\left(x-1\right)-\left(x+3\right)\left(x-2\right)-19$$

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

$$x^{3}-3x^{2}+3x-1=x^{3}-x^{2}-\left(x+3\right)\left(x-2\right)-19$$

Use the distributive property to multiply x+3 by x-2 and combine like terms.

$$x^{3}-3x^{2}+3x-1=x^{3}-x^{2}-\left(x^{2}+x-6\right)-19$$

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

$$x^{3}-3x^{2}+3x-1=x^{3}-x^{2}-x^{2}-x+6-19$$

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

$$x^{3}-3x^{2}+3x-1=x^{3}-2x^{2}-x+6-19$$

Subtract 19 from 6 to get -13.

$$x^{3}-3x^{2}+3x-1=x^{3}-2x^{2}-x-13$$

Subtract x^{3} from both sides.

$$x^{3}-3x^{2}+3x-1-x^{3}=-2x^{2}-x-13$$

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

$$-3x^{2}+3x-1=-2x^{2}-x-13$$

Add 2x^{2} to both sides.

$$-3x^{2}+3x-1+2x^{2}=-x-13$$

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

$$-x^{2}+3x-1=-x-13$$

Add x to both sides.

$$-x^{2}+3x-1+x=-13$$

Combine 3x and x to get 4x.

$$-x^{2}+4x-1=-13$$

Add 1 to both sides.

$$-x^{2}+4x=-13+1$$

Add -13 and 1 to get -12.

$$-x^{2}+4x=-12$$

Divide both sides by -1.

$$\frac{-x^{2}+4x}{-1}=-\frac{12}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{4}{-1}x=-\frac{12}{-1}$$

Divide 4 by -1.

$$x^{2}-4x=-\frac{12}{-1}$$

Divide -12 by -1.

$$x^{2}-4x=12$$

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

$$x^{2}-4x+\left(-2\right)^{2}=12+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=12+4$$

Add 12 to 4.

$$x^{2}-4x+4=16$$

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}.

$$\left(x-2\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{16}$$

Simplify.

$$x-2=4 x-2=-4$$

Add 2 to both sides of the equation.

$$x=6 x=-2$$