What if only one of the terms is a perfect cube?

Sum of Cubes requires both terms to be perfect cubes. If only one is, you cannot use this formula. Try factoring by grouping, finding a common factor, or another factoring technique instead.

Why cannot the quadratic factor be factored further?

The quadratic a^2 - ab + b^2 from the Sum of Cubes formula has a negative discriminant, meaning it has no real roots. This is why it cannot be factored into real linear factors.

Is Sum of Cubes the only way to factor a sum of cubes?

For a binomial of two perfect cubes, Sum of Cubes is the primary and most direct method. In some cases a common factor may first be present, but once removed, Sum of Cubes handles the remaining expression.

How do I tell Sum of Cubes apart from Difference of Cubes?

Sum of Cubes handles a^3 + b^3 (addition) and gives (a+b)(a^2 - ab + b^2). Difference of Cubes handles a^3 - b^3 (subtraction) and gives (a-b)(a^2 + ab + b^2). The key difference is the operation (plus or minus) and the sign of the middle term in the quadratic.

Sum of Cubes

Sum of Cubes is a factoring formula that breaks a binomial with two cubic terms into a product of a linear and a quadratic factor.

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

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What each symbol means

What Sum of Cubes takes

Symbol	Meaning
$a$	The first term in the binomial; it must be a perfect cube, meaning it equals some value raised to the third power.
$b$	The second term in the binomial; it must also be a perfect cube, and it is added to $a^3$ rather than subtracted.

When to use it

When you see two perfect cubes being added together, use Sum of Cubes to factor them instantly.

Level

Usually taught in: Algebra II

Worked examples

1. Factor a simple sum of cubes

Problem

Factor

x^3 + 27

$x^3 + 27 = x^3 + 3^3$
Recognize that $27 = 3^3$ , so this is a sum of two perfect cubes.
$x^3 + 3^3 = (x + 3)(x^2 - (x)(3) + 3^2)$
Apply the Sum of Cubes formula $(a)^3 + (b)^3 = (a+b)(a^2 - ab + b^2)$ with $a = x$ and $b = 3$ .
$(x + 3)(x^2 - 3x + 9)$
Simplify by evaluating $-(x)(3) = -3x$ and $3^2 = 9$ .

Answer: $(x + 3)(x^2 - 3x + 9)$

Sum of Cubes directly factors this binomial. Since both $x^3$ and $27$ are perfect cubes, we apply the formula and are done.

2. Factor a sum of cubes with coefficients

Problem

Factor

64a^3 + 1

$64a^3 + 1 = (4a)^3 + 1^3$
Recognize that $64a^3 = (4a)^3$ and $1 = 1^3$ , so we have a sum of perfect cubes.
$(4a)^3 + 1^3 = (4a + 1)((4a)^2 - (4a)(1) + 1^2)$
Apply the Sum of Cubes formula with $a = 4a$ and $b = 1$ .
$(4a + 1)(16a^2 - 4a + 1)$
Evaluate $(4a)^2 = 16a^2$ , $(4a)(1) = 4a$ , and $1^2 = 1$ to get the final factorization.

Answer: $(4a + 1)(16a^2 - 4a + 1)$

Recognizing that $64a^3 = (4a)^3$ is the key step. Once both terms are identified as perfect cubes, the formula applies directly.

3. Factor a sum of cubes from a storage capacity problem

Problem

A cloud storage service offers two account types. One stores

m^3

gigabytes, the other stores

125

gigabytes. The total storage is

m^3 + 125

gigabytes. Factor this expression to analyze the relationship between account types.

$m^3 + 125 = m^3 + 5^3$
Recognize that $125 = 5^3$ , so this is a sum of two perfect cubes.
$m^3 + 5^3 = (m + 5)(m^2 - (m)(5) + 5^2)$
Apply the Sum of Cubes formula with $a = m$ and $b = 5$ .
$(m + 5)(m^2 - 5m + 25)$
Simplify by computing $-(m)(5) = -5m$ and $5^2 = 25$ .

Answer: $(m + 5)(m^2 - 5m + 25)$

The factorization shows how total storage is a product of a linear factor and a quadratic factor, revealing the capacity relationship structure.

Common mistakes

Where Sum of Cubes usually goes wrong

The mistake	Why it is wrong	The fix
Writing $(x + 2)(x^2 - 2x - 4)$ when factoring $x^3 + 8$ .	The sign pattern in the quadratic factor is often confused; the middle term must be negative and the last term must be positive.	The correct factorization is $(x + 2)(x^2 - 2x + 4)$ where the middle term is $-ab = -2x$ and the last term is $+b^2 = +4$ .
Using $(a + b)(a^2 + ab + b^2)$ to factor $a^3 + b^3$ .	The Sum of Cubes formula is confused with the Difference of Cubes formula, which has a positive middle term instead of negative.	For Sum of Cubes, use $(a+b)(a^2 - ab + b^2)$ . For Difference of Cubes $a^3 - b^3$ , use $(a-b)(a^2 + ab + b^2)$ with a positive middle term.
Trying to factor $x^3 + y^3$ as $(x + y)^3$ or into two identical binomials.	Sum of Cubes produces one linear factor and one quadratic factor, not two identical factors or a perfect cube.	The correct factorization is $(x + y)(x^2 - xy + y^2)$ , and the quadratic factor cannot be factored further over the real numbers.

Tips and when to use something else

Always verify that both terms are perfect cubes before applying the formula. For instance, $x^3 + 10$ cannot use Sum of Cubes since $10$ is not a perfect cube.
Remember the sign pattern: $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ has a negative middle term. If you see $a^3 - b^3$ instead, you need Difference of Cubes $(a-b)(a^2 + ab + b^2)$ with a positive middle term.
The quadratic factor in the result is typically irreducible over the reals, so your factorization is complete. Do not attempt to factor it further.
Sum of Cubes applies only to addition; subtraction requires a different method. Always check whether you have a sum or difference before selecting your factoring strategy.

Related formulas

Difference of CubesDifference of Cubes is a factoring pattern that breaks down any cubic binomial subtraction into a linear and a quadratic factor.Factoring by GroupingFactoring by grouping factors four-term polynomials by pairing terms and extracting common factors—use it when you cannot factor out a GCF from all terms.Distributive PropertyThe Distributive Property lets you multiply a number by a sum or difference by distributing the multiplication to each term separately.

Frequently asked questions

What if only one of the terms is a perfect cube?: Sum of Cubes requires both terms to be perfect cubes. If only one is, you cannot use this formula. Try factoring by grouping, finding a common factor, or another factoring technique instead.
Why cannot the quadratic factor be factored further?: The quadratic $a^2 - ab + b^2$ from the Sum of Cubes formula has a negative discriminant, meaning it has no real roots. This is why it cannot be factored into real linear factors.
Is Sum of Cubes the only way to factor a sum of cubes?: For a binomial of two perfect cubes, Sum of Cubes is the primary and most direct method. In some cases a common factor may first be present, but once removed, Sum of Cubes handles the remaining expression.
How do I tell Sum of Cubes apart from Difference of Cubes?: Sum of Cubes handles $a^3 + b^3$ (addition) and gives $(a+b)(a^2 - ab + b^2)$ . Difference of Cubes handles $a^3 - b^3$ (subtraction) and gives $(a-b)(a^2 + ab + b^2)$ . The key difference is the operation (plus or minus) and the sign of the middle term in the quadratic.

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Reviewed 2026-09-18