Sum and Difference of Cubes

Why a sum or difference of cubes factors

Try a specific case first: $x^3 + 8$. Since $8 = 2^3$, both terms are perfect cubes, and the claim is that this factors as $(x + 2)(x^2 - 2x + 4)$. Check that claim the way you check any proposed factoring, by expanding the product:

The $-2x^2$ and $+2x^2$ cancel, and so do $+4x$ and $-4x$, leaving $x^3 + 8$. Nothing about $x$ and $2$ was special there; the same cancellation happens for any two terms $a$ and $b$, which gives two general identities, one for each sign:

Each is a binomial times a trinomial, and neither asks to be taken on faith. The proof below confirms both in general, the same way you already 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.

Nothing about that cancellation is arbitrary, and it is also what forces every sign in the two identities, 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 sign before the trinomial’s last term, $b^2$, is always plus, in both identities.

These are not three rules to accept blindly; the proof above forces all three. Same: the sign on $b$ decides whether the product ends in $+b^3$ or $-b^3$. Opposite: that sign on the middle term is exactly what makes the four cross terms cancel instead of piling up. Always Positive: that sign work all happens on the binomial’s $b$, so the trinomial’s last term keeps a plain plus in both identities.

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 whole number, or a term with a whole-number coefficient, that is something cubed. The perfect cubes from $1$ through $10$ are worth knowing on sight:

Variable terms can be cubes too, but a term needs two things to qualify: its coefficient must be a perfect cube, and every variable’s exponent must be a multiple of $3$ (since cubing multiplies an exponent by $3$). A bare power like $x^3$ is $(x)^3$, and $8x^3 = (2x)^3$, $27x^3 = (3x)^3$, and $64x^3 = (4x)^3$ all pass both tests. But $8x^2$ is not a perfect cube, even though $8$ is: the exponent $2$ is not a multiple of $3$. This lesson sticks to that plain case, exponent $3$, so $a$ is always a single variable term like $x$ or $2x$. 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; try it on $x^2 + 4x + 16$ and see what happens.

To factor $x^2 + 4x + 16$ into $(x + p)(x + q)$ you would need two integers whose product is $16$ and whose sum is $4$. Since both $16$ and $4$ are positive, $p$ and $q$ must both be positive too (two negatives would multiply to a positive but add to a negative). The positive 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.

That is the pattern for every sum or difference of cubes in this lesson: once you have removed any common factor and applied the cube identity, you have the final, complete factorization with integer coefficients.

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