# ##\$ Difference of Cubes

The Difference of Cubes Formula is a shortcut that you can use anytime you need to factor an expression that has a perfect cube subtracted from another perfect cube.

If you have two perfect cubes added instead of subtracted, you can use the Sum of Cubes Formula.

## How to Factor a Difference of Cubes

1. Make sure the polynomial is a difference of cubes.
• Does the polynomial have two terms connected with an subtraction sign?
• Are the terms perfect cubes?
2. Find the cube root of each term.
3. Plug the results of Step 2 into the Difference of Cubes Formula.

## Examples

Factor: $x^3-27$

Is $$\yellow x^3-27$$ a difference of cubes?

Yes, $$\yellow x^3$$ and $$\yellow 27$$ are perfect cubes that are connected with a subtraction sign.

What are the cube roots of $$\yellow x^3$$ and $$\yellow 27$$?

$\sqrt[3]{\yellow x^3}={\yellow x}$

$\sqrt[3]{\yellow 27}={\yellow 3}$

What is the factored form of $$\yellow x^3-27$$?

To find the factored form of $$\yellow x^3-27$$, I will substitute $$\yellow x$$ and $$\yellow 3$$ into the Difference of Cubes Formula.

$x^3-27=({\yellow x}-{\yellow 3})({\yellow x}^2+{\yellow 3}{\yellow x}+{\yellow 3}^2)$

When I simplify the $${\yellow 3}^2$$ term I get…

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

The factored form of $$\yellow x^3-27$$ is…

$\yellow (x-3)(x^2+3x+9)$

Factor: $125y^3-64x^3$

Is $$\blue 125y^3-64x^3$$ a difference of cubes?

Yes, $$\blue 125y^3$$ and $$\blue 64x^3$$ are perfect cubes that are connected with a subtraction sign.

What are the cube roots of $$\blue 125y^3$$ and $$\blue 64x^3$$?

$\sqrt[3]{\blue 125y^3}={\blue 5y}$

$\sqrt[3]{\blue 64x^3}={\blue 4x}$

What is the factored form of $$\blue 125y^3-64x^3$$?

To find the factored form of $$\blue 125y^3-64x^3$$, I will substitute $$\blue 5y$$ and $$\blue 4x$$ into the Difference of Cubes Formula.

$125y^3-64x^3=({\blue 5y}-{\blue 4x})(({\blue 5y})^2+({\blue 4x})({\blue 5y})+({\blue 4x})^2)$

When I simplify all the terms I get…

$125y^3-64x^3=({\blue 5y}-{\blue 4x})({\blue 25y^2}+{\blue 20xy}+{\blue 16x^2})$

The factored form of $$\blue 125y^3-64x^3$$ is…

$\blue (5y-4x)(25y^2+20xy+16x^2)$

Factor: $27x^{15}-8y^6$

Is $$\green 27x^{15}-8y^6$$ a difference of cubes?

Yes, $$\green 27x^{15}$$ and $$\green 8y^6$$ are perfect cubes that are connected with an subtraction sign.

What are the cube roots of $$\green 27x^{15}$$ and $$\green 8y^6$$?

$\sqrt[3]{\green 27x^{15}}={\green 3x^5}$

$\sqrt[3]{\green 8y^6}={\green 2y^2}$

What is the factored form of $$\green 27x^{15}-8y^6$$?

To find the factored form of $$\green 27x^{15}-8y^6$$, I will substitute $$\green 3x^5$$ and $$\green 2y^2$$ into the Difference of Cubes Formula.

$27x^{15}-8y^6=({\green 3x^5}-{\green 2y^2})(({\green 3x^5})^2+({\green 2y^2})({\green 3x^5})+({\green 2y^2})^2)$

When I simplify all the terms I get…

$27x^{15}-8y^6=({\green 3x^5}-{\green 2y^2})({\green 9x^{10}}+{\green 6x^5y^2}+{\green 4y^4})$

The factored form of $$\green 27x^{15}-8y^6$$ is…

$\green (3x^5-2y^2)(9x^{10}+6x^5y^2+4y^4)$

## Why It Works

The Difference of Cubes Formula works because factoring “un-does” polynomial multiplication.

You can see where the formula comes from if you reverse engineer the process and multiply $$(a-b)$$ and $$(a^2+ab+b^2)$$.

To understand where the formula comes from…

1. Multiply $$(a-b)(a^2+ab+b^2)$$.
2. Undo the multiplication to find the Difference of Cubes Formula.

I like using the multiplication algorithm to multiply polynomials, but you can also use the box method.

$\begin {array}{cccc} & a^2 & +ab & +b^2 \\ \times & & {\yellow a} & {\green -b} \\ \hline & {\green -a^2b} & {\green -ab^2} & {\green -b^3} \\ {\yellow a^3} & {\yellow +a^2b} & {\yellow +ab^2} & \\ \hline {\yellow a^3} & 0 & 0 & {\green -b^3} \end{array}$

Multiplying the polynomials gives us a sum of cubes…

$(a-b)(a^2+ab+b^2)=a^3-b^3$

That means we can “un-do” the multiplication and write the equation backwards to find the Difference of Cubes Formula…

$a^3-b^3=(a-b)(a^2+ab+b^2)$