The Sum of Cubes Formula is a shortcut that you can use anytime you need to factor an expression that has two perfect cubes added to each other.
If you have two perfect cubes subtracted instead of added, you can use the Difference of Cubes Formula.
Sum of Cubes Formula
\[a^3+b^3=(a+b)(a^2-ab+b^2)\]
How to Factor a Sum of Cubes
- Make sure the polynomial is a sum of cubes.
- Does the polynomial have two terms connected with an addition sign?
- Are the terms perfect cubes?
- Find the cube root of each term.
- Plug the results of Step 2 into the Sum of Cubes Formula.
Examples
Factor: \[x^3+8\]
Is \(\yellow x^3+8\) a sum of cubes?
Yes, \(\yellow x^3\) and \(\yellow 8\) are perfect cubes that are connected with an addition sign.
What are the cube roots of \(\yellow x^3\) and \(\yellow 8\)?
\[\sqrt[3]{\yellow x^3}={\yellow x}\]
\[\sqrt[3]{\yellow 8}={\yellow 2}\]
What is the factored form of \(\yellow x^3+8\)?
To find the factored form of \(\yellow x^3+8\), I will substitute \(\yellow x\) and \(\yellow 2\) into the Sum of Cubes Formula.
\[x^3+8=({\yellow x}+{\yellow 2})({\yellow x}^2-{\yellow 2}{\yellow x}+{\yellow 2}^2)\]
When I simplify the \({\yellow 2}^2\) term I get…
\[x^3+8=({\yellow x}+{\yellow 2})({\yellow x}^2-{\yellow 2}{\yellow x}+{\yellow 4})\]
The factored form of \(\yellow x^3+8\) is…
\[\yellow (x+2)(x^2-2x+4)\]
Factor: \[27y^3+64x^3\]
Is \(\blue 27y^3+64x^3\) a sum of cubes?
Yes, \(\blue 27y^3\) and \(\blue 64x^3\) are perfect cubes that are connected with an addition sign.
What are the cube roots of \(\blue 27y^3\) and \(\blue 64x^3\)?
\[\sqrt[3]{\blue 27y^3}={\blue 3y}\]
\[\sqrt[3]{\blue 64x^3}={\blue 4x}\]
What is the factored form of \(\blue 27y^3+64x^3\)?
To find the factored form of \(\blue 27y^3+64x^3\), I will substitute \(\blue 3y\) and \(\blue 4x\) into the Sum of Cubes Formula.
\[27y^3+64x^3=({\blue 3y}+{\blue 4x})(({\blue 3y})^2-({\blue 4x})({\blue 3y})+({\blue 4x})^2)\]
When I simplify all the terms I get…
\[27y^3+64x^3=({\blue 3y}+{\blue 4x})({\blue 9y^2}-{\blue 12xy}+{\blue 16x^2})\]
The factored form of \(\blue 27y^3+64x^3\) is…
\[\blue (3y+4x)(9y^2-12xy+16x^2)\]
Factor: \[125x^{12}+8y^9\]
Is \(\green 125x^{12}+8y^9\) a sum of cubes?
Yes, \(\green 125x^{12}\) and \(\green 8y^9\) are perfect cubes that are connected with an addition sign.
What are the cube roots of \(\green 125x^{12}\) and \(\green 8y^9\)?
\[\sqrt[3]{\green 125x^{12}}={\green 5x^4}\]
\[\sqrt[3]{\green 8y^9}={\green 2y^3}\]
What is the factored form of \(\green 125x^{12}+8y^9\)?
To find the factored form of \(\green 125x^{12}+8y^9\), I will substitute \(\green 5x^4\) and \(\green 2y^3\) into the Sum of Cubes Formula.
\[125x^{12}+8y^9=({\green 5x^4}+{\green 2y^3})(({\green 5x^4})^2-({\green 2y^3})({\green 5x^4})+({\green 2y^3})^2)\]
When I simplify all the terms I get…
\[125x^{12}+8y^9=({\green 5x^4}+{\green 2y^3})({\green 25x^8}-{\green 10x^4y^3}+{\green 4y^6})\]
The factored form of \(\green 125x^{12}+8y^9\) is…
\[\green (5x^4+2y^3)(25x^8-10x^4y^3+4y^6)\]
Why It Works
The Sum 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 Sum 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 Sum of Cubes Formula…
\[a^3+b^3=(a+b)(a^2-ab+b^2)\]
