Solve x(x-3)^2-(x-3)^3=3(x-3)

$$x ( x - 3 ) ^ { 2 } - ( x - 3 ) ^ { 3 } = 3 ( x - 3 )$$

Solve for x

$$x=3$$

$$x=4$$

Steps Using Factoring By Grouping

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

$$x\left(x^{2}-6x+9\right)-\left(x-3\right)^{3}=3\left(x-3\right)$$

Use the distributive property to multiply x by x^{2}-6x+9.

$$x^{3}-6x^{2}+9x-\left(x-3\right)^{3}=3\left(x-3\right)$$

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

$$x^{3}-6x^{2}+9x-\left(x^{3}-9x^{2}+27x-27\right)=3\left(x-3\right)$$

To find the opposite of x^{3}-9x^{2}+27x-27, find the opposite of each term.

$$x^{3}-6x^{2}+9x-x^{3}+9x^{2}-27x+27=3\left(x-3\right)$$

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

$$-6x^{2}+9x+9x^{2}-27x+27=3\left(x-3\right)$$

Combine -6x^{2} and 9x^{2} to get 3x^{2}.

$$3x^{2}+9x-27x+27=3\left(x-3\right)$$

Combine 9x and -27x to get -18x.

$$3x^{2}-18x+27=3\left(x-3\right)$$

Use the distributive property to multiply 3 by x-3.

$$3x^{2}-18x+27=3x-9$$

Subtract 3x from both sides.

$$3x^{2}-18x+27-3x=-9$$

Combine -18x and -3x to get -21x.

$$3x^{2}-21x+27=-9$$

Add 9 to both sides.

$$3x^{2}-21x+27+9=0$$

Add 27 and 9 to get 36.

$$3x^{2}-21x+36=0$$

Divide both sides by 3.

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

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.

$$a+b=-7 ab=1\times 12=12$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.

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

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -7.

$$a=-4 b=-3$$

Rewrite x^{2}-7x+12 as \left(x^{2}-4x\right)+\left(-3x+12\right).

$$\left(x^{2}-4x\right)+\left(-3x+12\right)$$

Factor out x in the first and -3 in the second group.

$$x\left(x-4\right)-3\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x-3\right)$$

To find equation solutions, solve x-4=0 and x-3=0.

$$x=4 x=3$$

Steps Using the Quadratic Formula

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

$$x\left(x^{2}-6x+9\right)-\left(x-3\right)^{3}=3\left(x-3\right)$$

Use the distributive property to multiply x by x^{2}-6x+9.

$$x^{3}-6x^{2}+9x-\left(x-3\right)^{3}=3\left(x-3\right)$$

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

$$x^{3}-6x^{2}+9x-\left(x^{3}-9x^{2}+27x-27\right)=3\left(x-3\right)$$

To find the opposite of x^{3}-9x^{2}+27x-27, find the opposite of each term.

$$x^{3}-6x^{2}+9x-x^{3}+9x^{2}-27x+27=3\left(x-3\right)$$

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

$$-6x^{2}+9x+9x^{2}-27x+27=3\left(x-3\right)$$

Combine -6x^{2} and 9x^{2} to get 3x^{2}.

$$3x^{2}+9x-27x+27=3\left(x-3\right)$$

Combine 9x and -27x to get -18x.

$$3x^{2}-18x+27=3\left(x-3\right)$$

Use the distributive property to multiply 3 by x-3.

$$3x^{2}-18x+27=3x-9$$

Subtract 3x from both sides.

$$3x^{2}-18x+27-3x=-9$$

Combine -18x and -3x to get -21x.

$$3x^{2}-21x+27=-9$$

Add 9 to both sides.

$$3x^{2}-21x+27+9=0$$

Add 27 and 9 to get 36.

$$3x^{2}-21x+36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -21 for b, and 36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 3\times 36}}{2\times 3}$$

Square -21.

$$x=\frac{-\left(-21\right)±\sqrt{441-4\times 3\times 36}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-21\right)±\sqrt{441-12\times 36}}{2\times 3}$$

Multiply -12 times 36.

$$x=\frac{-\left(-21\right)±\sqrt{441-432}}{2\times 3}$$

Add 441 to -432.

$$x=\frac{-\left(-21\right)±\sqrt{9}}{2\times 3}$$

Take the square root of 9.

$$x=\frac{-\left(-21\right)±3}{2\times 3}$$

The opposite of -21 is 21.

$$x=\frac{21±3}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{21±3}{6}$$

Now solve the equation x=\frac{21±3}{6} when ± is plus. Add 21 to 3.

$$x=\frac{24}{6}$$

Divide 24 by 6.

$$x=4$$

Now solve the equation x=\frac{21±3}{6} when ± is minus. Subtract 3 from 21.

$$x=\frac{18}{6}$$

Divide 18 by 6.

$$x=3$$

The equation is now solved.

$$x=4 x=3$$

Steps for Completing the Square

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

$$x\left(x^{2}-6x+9\right)-\left(x-3\right)^{3}=3\left(x-3\right)$$

Use the distributive property to multiply x by x^{2}-6x+9.

$$x^{3}-6x^{2}+9x-\left(x-3\right)^{3}=3\left(x-3\right)$$

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

$$x^{3}-6x^{2}+9x-\left(x^{3}-9x^{2}+27x-27\right)=3\left(x-3\right)$$

To find the opposite of x^{3}-9x^{2}+27x-27, find the opposite of each term.

$$x^{3}-6x^{2}+9x-x^{3}+9x^{2}-27x+27=3\left(x-3\right)$$

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

$$-6x^{2}+9x+9x^{2}-27x+27=3\left(x-3\right)$$

Combine -6x^{2} and 9x^{2} to get 3x^{2}.

$$3x^{2}+9x-27x+27=3\left(x-3\right)$$

Combine 9x and -27x to get -18x.

$$3x^{2}-18x+27=3\left(x-3\right)$$

Use the distributive property to multiply 3 by x-3.

$$3x^{2}-18x+27=3x-9$$

Subtract 3x from both sides.

$$3x^{2}-18x+27-3x=-9$$

Combine -18x and -3x to get -21x.

$$3x^{2}-21x+27=-9$$

Subtract 27 from both sides.

$$3x^{2}-21x=-9-27$$

Subtract 27 from -9 to get -36.

$$3x^{2}-21x=-36$$

Divide both sides by 3.

$$\frac{3x^{2}-21x}{3}=-\frac{36}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{21}{3}\right)x=-\frac{36}{3}$$

Divide -21 by 3.

$$x^{2}-7x=-\frac{36}{3}$$

Divide -36 by 3.

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

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=-12+\frac{49}{4}$$

Add -12 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{1}{4}$$

Factor x^{2}-7x+\frac{49}{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}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{1}{2} x-\frac{7}{2}=-\frac{1}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=4 x=3$$