Sum and Difference of Cubes

Why a sum or difference of cubes factors

There are two identities to establish, one for each sign:

Each is a binomial times a trinomial, and neither asks to be taken on faith. You can confirm both the same way you confirmed the difference of squares. Expand the right-hand side, multiplying every term of the binomial by every term of the trinomial, and see what survives.

The whole mechanism is those four middle terms falling away in two matched pairs. Nothing about it is arbitrary, and in particular the signs are forced, which is the next thing to pin down.

The SOAP sign pattern

The two identities differ only in where the signs go, and there is a tidy way to remember the pattern. It is often called SOAP, for Same, Opposite, Always Positive, read across the three signs you have to fill in:

• Same. The binomial takes the same sign as the original expression. A sum of cubes gives $(a + b)$; a difference gives $(a - b)$.
• Opposite. The middle term of the trinomial takes the opposite sign. A sum gives $a^2 - ab + b^2$; a difference gives $a^2 + ab + b^2$.
• Always Positive. The last term of the trinomial is $+b^2$, always positive, in both identities.

These are not three rules to accept blindly; the derivation above forces every one of them. Look back at the expansion of $(a + b)(a^2 - ab + b^2)$. The final $b^3$ came from multiplying $b$ by the trinomial’s last term $b^2$. So the sign you put in front of the binomial’s $b$ is exactly what decides whether the result ends in $+b^3$ or $-b^3$. That is the Same rule. The four middle terms cancelled only because the trinomial’s middle term carried the opposite sign from the binomial. Make both of those signs plus and the middle terms reinforce instead of cancelling, and no clean factorization survives. That is the Opposite rule. And the trinomial’s last term stayed $+b^2$ in both cases, because $b^2$ is a square and all of the sign work is handled by the binomial. That is Always Positive.

One warning the pattern makes visible: the middle term is a single $ab$, not $2ab$. The doubled middle term belongs to a perfect-square trinomial like $(a + b)^2 = a^2 + 2ab + b^2$, which is a different animal. The cube trinomial never doubles it.

Recognizing perfect cubes

Before you can use either identity you have to see two perfect cubes. A perfect cube is a number or term that is something cubed. The small perfect cubes are worth knowing on sight:

Variable terms can be cubes too. A bare power like $x^3$ is $(x)^3$, and a term with a coefficient is a cube when the coefficient is a perfect cube: $8x^3 = (2x)^3$, $27x^3 = (3x)^3$, and $64x^3 = (4x)^3$. To set up a factorization, take the cube root of each term to find $a$ and $b$. The cube root of $8x^3$ is $2x$, not $8x$, because you must cube both the $2$ and the $x$ to rebuild $8x^3$.

Factoring cubes that carry coefficients

Nothing in the pattern requires bare variables. Once you can take the cube root of each term, coefficient and all, you just read off $a$ and $b$ and let SOAP place the signs.

Why the trinomial does not factor further

After you factor a sum or difference of cubes, you are left with a trinomial like $x^2 + 4x + 16$ or $x^2 - 3x + 9$. It is tempting to keep going and try to factor that trinomial as well. Resist. For the cube pattern, that trinomial does not factor over the integers, and there is a concrete way to see why.

To factor $x^2 + 4x + 16$ into $(x + p)(x + q)$ you would need two integers whose product is $16$ and whose sum is $4$. The integer pairs that multiply to $16$ are $1$ and $16$, $2$ and $8$, and $4$ and $4$, which sum to $17$, $10$, and $8$. None of them sum to $4$, so no such factoring exists.

The same happens every time: the cube trinomial has no integer factor pair. That the trinomial has no real factors at all is confirmed by tools in a later chapter. When a factorization reaches the binomial-times-trinomial stage, you are finished, unless a common factor was hiding in front.

Pulling out a common factor first

An expression that is not a sum or difference of cubes as written can turn into one after you factor out a common factor. So always check for a common factor first. Take $2x^3 + 16$. Neither $2x^3$ nor $16$ is a perfect cube, so the pattern does not apply yet. But both terms are even, and once you pull out the $2$, what is left is a clean sum of cubes:

The habit is to factor out the greatest common factor first, then check whether what remains is a sum or difference of cubes. Skip that step and you either miss the cubes entirely or leave the answer half done.

Common mistakes