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.
Difference of Cubes Formula
\[a^3-b^3=(a-b)(a^2+ab+b^2)\]
How to Factor a Difference of Cubes
- 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?
- Find the cube root of each term.
- 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…
- Multiply \((a-b)(a^2+ab+b^2)\).
- 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…
\[a^3-b^3=(a-b)(a^2+ab+b^2)\]
