Solve p(p-3)^2-(p-3)^3=0

$$p ( p - 3 ) ^ { 2 } - ( p - 3 ) ^ { 3 } = 0$$

Solve for p

$$p=3$$

Steps Using Factoring By Grouping

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

$$p\left(p^{2}-6p+9\right)-\left(p-3\right)^{3}=0$$

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

$$p^{3}-6p^{2}+9p-\left(p-3\right)^{3}=0$$

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

$$p^{3}-6p^{2}+9p-\left(p^{3}-9p^{2}+27p-27\right)=0$$

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

$$p^{3}-6p^{2}+9p-p^{3}+9p^{2}-27p+27=0$$

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

$$-6p^{2}+9p+9p^{2}-27p+27=0$$

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

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

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

$$3p^{2}-18p+27=0$$

Divide both sides by 3.

$$p^{2}-6p+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as p^{2}+ap+bp+9. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\times 9=9$$

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 9.

$$-1,-9 -3,-3$$

Calculate the sum for each pair.

$$-1-9=-10 -3-3=-6$$

The solution is the pair that gives sum -6.

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

Rewrite p^{2}-6p+9 as \left(p^{2}-3p\right)+\left(-3p+9\right).

$$\left(p^{2}-3p\right)+\left(-3p+9\right)$$

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

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

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

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

Rewrite as a binomial square.

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

To find equation solution, solve p-3=0.

$$p=3$$

Steps Using the Quadratic Formula

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

$$p\left(p^{2}-6p+9\right)-\left(p-3\right)^{3}=0$$

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

$$p^{3}-6p^{2}+9p-\left(p-3\right)^{3}=0$$

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

$$p^{3}-6p^{2}+9p-\left(p^{3}-9p^{2}+27p-27\right)=0$$

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

$$p^{3}-6p^{2}+9p-p^{3}+9p^{2}-27p+27=0$$

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

$$-6p^{2}+9p+9p^{2}-27p+27=0$$

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

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

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

$$3p^{2}-18p+27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -18 for b, and 27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 3\times 27}}{2\times 3}$$

Square -18.

$$p=\frac{-\left(-18\right)±\sqrt{324-4\times 3\times 27}}{2\times 3}$$

Multiply -4 times 3.

$$p=\frac{-\left(-18\right)±\sqrt{324-12\times 27}}{2\times 3}$$

Multiply -12 times 27.

$$p=\frac{-\left(-18\right)±\sqrt{324-324}}{2\times 3}$$

Add 324 to -324.

$$p=\frac{-\left(-18\right)±\sqrt{0}}{2\times 3}$$

Take the square root of 0.

$$p=-\frac{-18}{2\times 3}$$

The opposite of -18 is 18.

$$p=\frac{18}{2\times 3}$$

Multiply 2 times 3.

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

Divide 18 by 6.

$$p=3$$

Steps for Completing the Square

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

$$p\left(p^{2}-6p+9\right)-\left(p-3\right)^{3}=0$$

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

$$p^{3}-6p^{2}+9p-\left(p-3\right)^{3}=0$$

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

$$p^{3}-6p^{2}+9p-\left(p^{3}-9p^{2}+27p-27\right)=0$$

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

$$p^{3}-6p^{2}+9p-p^{3}+9p^{2}-27p+27=0$$

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

$$-6p^{2}+9p+9p^{2}-27p+27=0$$

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

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

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

$$3p^{2}-18p+27=0$$

Subtract 27 from both sides. Anything subtracted from zero gives its negation.

$$3p^{2}-18p=-27$$

Divide both sides by 3.

$$\frac{3p^{2}-18p}{3}=-\frac{27}{3}$$

Dividing by 3 undoes the multiplication by 3.

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

Divide -18 by 3.

$$p^{2}-6p=-\frac{27}{3}$$

Divide -27 by 3.

$$p^{2}-6p=-9$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -3.

$$p^{2}-6p+9=-9+9$$

Add -9 to 9.

$$p^{2}-6p+9=0$$

Factor p^{2}-6p+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(p-3\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(p-3\right)^{2}}=\sqrt{0}$$

Simplify.

$$p-3=0 p-3=0$$

Add 3 to both sides of the equation.

$$p=3 p=3$$

The equation is now solved. Solutions are the same.

$$p=3$$