When do I use Difference of Cubes vs. Sum of Cubes?

Use Difference of Cubes (a^3 - b^3) when you have one cubic term being subtracted from another. Use Sum of Cubes (a^3 + b^3) when the cubic terms are being added. The factorization formulas are different: Difference of Cubes uses (a - b)(a^2 + ab + b^2), while Sum of Cubes uses (a + b)(a^2 - ab + b^2).

Can the quadratic factor from Difference of Cubes be factored further?

No, the quadratic a^2 + ab + b^2 cannot be factored further over the real numbers. Its discriminant is negative, meaning it has no real roots and cannot be written as a product of two linear factors with real coefficients.

What if my expression has a negative sign first, like -8 + x^3?

Rearrange to standard form by moving the positive cubic term first: x^3 - 8. Then apply the Difference of Cubes formula as usual to get (x - 2)(x^2 + 2x + 4).

Why does the middle term in the quadratic factor use addition, not subtraction?

The formula a^3 - b^3 = (a - b)(a^2 + ab + b^2) always has the middle term as +ab because that's what arises when you expand (a - b)(a^2 + ab + b^2) step by step. If you used subtraction (-ab), the expansion would not equal a^3 - b^3.

Difference of Cubes

Difference of Cubes is a factoring pattern that breaks down any cubic binomial subtraction into 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 Difference of Cubes takes

Symbol	Meaning
$a$	The first number or expression being cubed; it can be any real number or variable, and confusing it with $b$ will lead to an incorrect factorization.
$b$	The number or expression being subtracted after being cubed; it represents what you're subtracting, and swapping it with $a$ changes the sign of the first factor from $(a - b)$ to $(b - a)$ .

When to use it

Reach for Difference of Cubes when you need to factor an expression where one cubic term is subtracted from another.

Level

Usually taught in: Algebra II

Worked examples

1. Factor a simple cubic binomial with small integers

Problem

Factor

x^3 - 8

$x^3 - 8 = x^3 - 2^3$
Identify each term as a perfect cube; $8 = 2^3$ , so we have $a = x$ and $b = 2$ .
$(a - b)(a^2 + ab + b^2) = (x - 2)(x^2 + x \cdot 2 + 2^2)$
Substitute $a = x$ and $b = 2$ into the Difference of Cubes formula.
$(x - 2)(x^2 + 2x + 4)$
Evaluate $2^2 = 4$ and simplify $x \cdot 2 = 2x$ to get the final factorization.

Answer: $(x - 2)(x^2 + 2x + 4)$

This problem uses the Difference of Cubes formula because we have two perfect cubes being subtracted. The pattern fits directly: $x^3 - 2^3$ factors into a linear term $(x - 2)$ and a quadratic term $(x^2 + 2x + 4)$ .

2. Factor a cubic binomial with composite variables

Problem

Factor

1000 - x^3y^3

$1000 - x^3y^3 = 10^3 - (xy)^3$
Identify $1000 = 10^3$ and $x^3y^3 = (xy)^3$ , so this is a difference of cubes with $a = 10$ and $b = xy$ .
$(10 - xy)(10^2 + 10 \cdot xy + (xy)^2)$
Substitute $a = 10$ and $b = xy$ into the Difference of Cubes formula.
$(10 - xy)(100 + 10xy + x^2y^2)$
Evaluate $10^2 = 100$ and $(xy)^2 = x^2y^2$ ; the middle term is $10 \cdot xy = 10xy$ .

Answer: $(10 - xy)(100 + 10xy + x^2y^2)$

This example requires more care because the second cubic term contains multiple variables. Recognizing that $x^3y^3 = (xy)^3$ allows us to apply the formula correctly; the key insight is treating $xy$ as a single unit for the purpose of the formula.

3. Factor a cubic binomial in a road-trip cargo problem

Problem

A delivery truck travels a route with two legs. The first leg of the journey requires a cargo volume of

(3x)^3

cubic feet (where

x

is a distance-based scaling factor), and the second leg requires only

8

cubic feet of cargo space. To calculate the difference in cargo volume needs, factor the expression

(3x)^3 - 8

$(3x)^3 - 8 = (3x)^3 - 2^3$
Identify $8 = 2^3$ , so this is a Difference of Cubes with $a = 3x$ and $b = 2$ .
$(3x - 2)((3x)^2 + 3x \cdot 2 + 2^2)$
Substitute $a = 3x$ and $b = 2$ into the Difference of Cubes formula.
$(3x - 2)(9x^2 + 6x + 4)$
Evaluate $(3x)^2 = 9x^2$ , $3x \cdot 2 = 6x$ , and $2^2 = 4$ to get the completely factored form.

Answer: $(3x - 2)(9x^2 + 6x + 4)$

This real-world scenario shows how Difference of Cubes applies when coefficients and variables appear in the cubic term. The factored form $(3x - 2)(9x^2 + 6x + 4)$ represents the decomposition of the cargo volume difference in a simplified way that could be useful for further calculations or analysis.

Common mistakes

Where Difference of Cubes usually goes wrong

The mistake	Why it is wrong	The fix
Writing $x^3 - 27 = (x - 3)(x^2 - 6x + 9)$ by using subtraction in the quadratic instead of addition.	The middle term in the quadratic factor must always be $ab$ with the same sign as in the formula, not $-ab$ . This mistake often comes from confusing Difference of Cubes with $(a - b)^3$ or misremembering the formula.	The correct factorization is $(x - 3)(x^2 + 3x + 9)$ — use addition ( $ab$ ) in the quadratic, not subtraction ( $-ab$ ).
Writing $x^3 - 8 = (x - 2)$ and stopping, thinking the factorization is complete.	The Difference of Cubes formula always produces two factors: a linear factor $(a - b)$ and a quadratic factor $(a^2 + ab + b^2)$ . Omitting either factor leaves the expression incomplete.	The complete factorization must be $(x - 2)(x^2 + 2x + 4)$ ; the quadratic factor cannot be ignored or simplified away.
Factoring $x^3 - 8$ using the Sum of Cubes pattern and writing $(x - 2)(x^2 - 2x + 4)$ with negative signs instead of positive.	The Sum of Cubes formula $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ is different from Difference of Cubes; using the wrong formula produces an incorrect middle term $-ab$ instead of $ab$ .	For Difference of Cubes, use $(a - b)(a^2 + ab + b^2)$ with positive signs in the quadratic factor: $(x - 2)(x^2 + 2x + 4)$ .

Tips and when to use something else

The quadratic factor $a^2 + ab + b^2$ never factors further over the real numbers (its discriminant is negative), so you can always stop once you apply the formula.
You can check your answer by expanding the factored form using FOIL or distribution; if you get back the original expression, you factored correctly.
When you see $a^3 - b^3$ , think Difference of Cubes immediately; but if you see $a^3 + b^3$ instead, switch to the Sum of Cubes formula $(a + b)(a^2 - ab + b^2)$ — the signs in the second factor change.
Don't forget to look for a Greatest Common Factor first — if your cubic binomial has one, factor it out before applying the Difference of Cubes formula.

Related formulas

Sum of CubesSum of Cubes is a factoring formula that breaks a binomial with two cubic terms into a product of a linear and a quadratic factor.Difference of SquaresInstantly factor binomials that are differences of perfect squares by recognizing the pattern and splitting into a product of a sum and difference.Factoring TrinomialsBreak a trinomial into two binomials by finding two numbers that add to b and multiply to c; use when solving equations or simplifying.Greatest Common Factor of a PolynomialFactor a polynomial by identifying and pulling out the single largest expression that divides evenly into every term in the polynomial.

Frequently asked questions

When do I use Difference of Cubes vs. Sum of Cubes?: Use Difference of Cubes ( $a^3 - b^3$ ) when you have one cubic term being subtracted from another. Use Sum of Cubes ( $a^3 + b^3$ ) when the cubic terms are being added. The factorization formulas are different: Difference of Cubes uses $(a - b)(a^2 + ab + b^2)$ , while Sum of Cubes uses $(a + b)(a^2 - ab + b^2)$ .
Can the quadratic factor from Difference of Cubes be factored further?: No, the quadratic $a^2 + ab + b^2$ cannot be factored further over the real numbers. Its discriminant is negative, meaning it has no real roots and cannot be written as a product of two linear factors with real coefficients.
What if my expression has a negative sign first, like $-8 + x^3$ ?: Rearrange to standard form by moving the positive cubic term first: $x^3 - 8$ . Then apply the Difference of Cubes formula as usual to get $(x - 2)(x^2 + 2x + 4)$ .
Why does the middle term in the quadratic factor use addition, not subtraction?: The formula $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ always has the middle term as $+ab$ because that's what arises when you expand $(a - b)(a^2 + ab + b^2)$ step by step. If you used subtraction ( $-ab$ ), the expansion would not equal $a^3 - b^3$ .

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Reviewed 2026-09-18