Cube of a Binomial - Formula & Worked Examples

What Is The Cube of a Binomial?

The cube of a binomial is the result of raising a two-term expression to the third power, and it expands to four terms: $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. The expansion is not $a^3 + b^3$; the two middle terms $3a^2b$ and $3ab^2$ are always there, and dropping them is the single most common error.

The two standard forms are:

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

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

Notice the only difference: in $(a-b)^3$ the signs alternate, plus-minus-plus-minus. The coefficients $1, 3, 3, 1$ are identical, and they match the fourth row of Pascal's triangle — the same pattern the binomial theorem produces for any power.

Variable Glossary:

How Is The Cube of a Binomial Formula Derived?

You derive it by multiplying step by step, never skipping a line. Start by writing the cube as a square times one more factor:

$$(a+b)^3 = (a+b)^2 (a+b)$$

First expand the square:

$$(a+b)^2 = a^2 + 2ab + b^2$$

Now multiply that trinomial by $(a+b)$, distributing each term:

$$a^2(a+b) = a^3 + a^2b$$

$$2ab(a+b) = 2a^2b + 2ab^2$$

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

Add the three lines and combine like terms. The $a^2b$ terms total $a^2b + 2a^2b = 3a^2b$, and the $ab^2$ terms total $2ab^2 + ab^2 = 3ab^2$:

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

The coefficient 3 is not arbitrary. It counts how many of the eight blocks in the cube share the same dimensions — there are exactly three slabs measuring $a \times a \times b$ and three measuring $a \times b \times b$.

Examples of Cube of a Binomial

Example 1

Expand $(x + 2)^3$.

Match to $(a+b)^3$ with $a = x$ and $b = 2$:

$$x^3 + 3(x^2)(2) + 3(x)(2^2) + 2^3$$

$$= x^3 + 6x^2 + 12x + 8$$

Final answer: $x^3 + 6x^2 + 12x + 8$.

Example 2

Expand $(y - 3)^3$.

This is where the most common mistake lives, so watch it happen first.

Wrong attempt. A student remembers "cube the terms" and writes $(y-3)^3 = y^3 - 27$, treating it like $a^3 - b^3$. Test it with a number: at $y = 5$, the true value is $(5-3)^3 = 2^3 = 8$, but $y^3 - 27 = 125 - 27 = 98$. Those are nowhere near each other, so the two middle terms cannot have been zero.

Correct. Use $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ with $a = y$, $b = 3$:

$$y^3 - 3(y^2)(3) + 3(y)(3^2) - 3^3$$

$$= y^3 - 9y^2 + 27y - 27$$

Check at $y = 5$: $125 - 225 + 135 - 27 = 8$. It matches.

Final answer: $y^3 - 9y^2 + 27y - 27$.

Example 3

Expand $(2a + 1)^3$ — a coefficient on the first term.

With first term $2a$ and second term $1$:

$$(2a)^3 + 3(2a)^2(1) + 3(2a)(1)^2 + 1^3$$

$$= 8a^3 + 3(4a^2)(1) + 3(2a)(1) + 1$$

$$= 8a^3 + 12a^2 + 6a + 1$$

Final answer: $8a^3 + 12a^2 + 6a + 1$.

Example 4

Expand $(3x - 2y)^3$ — two variables.

With $a = 3x$ and $b = 2y$, using the minus form:

$$(3x)^3 - 3(3x)^2(2y) + 3(3x)(2y)^2 - (2y)^3$$

$$= 27x^3 - 3(9x^2)(2y) + 3(3x)(4y^2) - 8y^3$$

$$= 27x^3 - 54x^2y + 36xy^2 - 8y^3$$

Final answer: $27x^3 - 54x^2y + 36xy^2 - 8y^3$.

Example 5

Use the cube of a binomial to evaluate $101^3$ mentally.

Write $101$ as $100 + 1$ and apply $(a+b)^3$ with $a = 100$, $b = 1$:

$$100^3 + 3(100)^2(1) + 3(100)(1)^2 + 1^3$$

$$= 1{,}000{,}000 + 30{,}000 + 300 + 1$$

$$= 1{,}030{,}301$$

Final answer: $101^3 = 1{,}030{,}301$.

Example 6

Expand $\left(x + \tfrac{1}{2}\right)^3$ — with a fraction.

With $a = x$, $b = \tfrac{1}{2}$:

$$x^3 + 3(x^2)\left(\tfrac{1}{2}\right) + 3(x)\left(\tfrac{1}{2}\right)^2 + \left(\tfrac{1}{2}\right)^3$$

$$= x^3 + \tfrac{3}{2}x^2 + 3(x)\left(\tfrac{1}{4}\right) + \tfrac{1}{8}$$

$$= x^3 + \tfrac{3}{2}x^2 + \tfrac{3}{4}x + \tfrac{1}{8}$$

Final answer: $x^3 + \tfrac{3}{2}x^2 + \tfrac{3}{4}x + \tfrac{1}{8}$.

How Does This Relate To The Sum And Difference of Cubes?

The cube of a binomial expands a bracket forward. Its close cousin, the sum and difference of cubes, runs the idea in reverse — it factors a two-term cubic expression $a^3 \pm b^3$ back into a product. The two identities are:

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

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

These are not the same as $(a+b)^3$ or $(a-b)^3$. The expansion $(a+b)^3$ has four terms; the sum of cubes $a^3 + b^3$ has only two and factors into a linear bracket times a quadratic. The link is that both come from the same multiplication structure — and a quick way to confirm the sum-of-cubes identity is to multiply the right side back out:

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

Every middle term cancels in pairs, leaving the two cubes. Recognising which form you are looking at — a cube to expand or a sum/difference of cubes to factor — is the first decision in any problem that involves cubed terms.

Why This Expansion Shows Up Beyond The Classroom

The cube of a binomial is the WHY behind any calculation that asks how a quantity raised to the third power changes when its base shifts a little. Think of the volume of a cube-shaped tank whose side is being increased from $a$ to $a+b$ — the extra volume is not $b^3$, it is the three slabs and the small cube together, $3a^2b + 3ab^2 + b^3$. Engineers sizing the extra material needed to enlarge a cubic container are computing exactly this difference.

It also seeds the binomial theorem, the general machine for raising a binomial to any power, and it connects to the perfect cube idea — a number or expression that is some quantity cubed. The factored mirror of this topic, the sum and difference of cubes, builds directly on recognising the $a^3 \pm b^3$ pattern.

Where Students Trip Up on The Cube of A Binomial

Mistake 1: Cubing each term and stopping

Where it slips in: Right at the start, when a student maps $(a+b)^3$ onto the (false) pattern $a^3 + b^3$ and never writes the middle terms.

Don't do this: Writing $(y - 3)^3 = y^3 - 27$.

The correct way: A binomial cube has four terms, not two. The first instinct on $(a+b)^3$ is to drop the cross-terms $3a^2b$ and $3ab^2$ — those middle terms are exactly what the formula exists to capture, and a quick numerical check (plug in a value) catches the error instantly.

Mistake 2: Forgetting the coefficient 3

Where it slips in: A student remembers there are middle terms but writes them as $a^2b$ and $ab^2$ without the factor of 3.

Don't do this: Writing $(x+2)^3 = x^3 + x^2(2) + x(4) + 8 = x^3 + 2x^2 + 4x + 8$.

The correct way: Each cross-term carries a coefficient of 3, because three of the eight blocks in the cube share those dimensions. The habit that fixes this is deriving the formula once by hand — students who expand $(a+b)^2(a+b)$ themselves see the 3 arrive, and stop guessing the coefficient afterward.

Mistake 3: Mishandling the signs in $(a-b)^3$

Where it slips in: With a subtraction inside the binomial, students apply the plus-form coefficients but forget that the signs alternate.

Don't do this: Writing $(3x - 2y)^3 = 27x^3 - 54x^2y - 36xy^2 - 8y^3$ — the third term should be positive.

The correct way: In $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$, the signs go plus, minus, plus, minus. The $+3ab^2$ term is positive because $b$ appears to an even power there.

Practice Questions on Cube of a Binomial

Expand each fully, watching the coefficient 3 and the signs. Where you can, check by plugging in a number. Answers follow.

1. Expand $(x + 3)^3$.

2. Expand $(a - 4)^3$.

3. Expand $(2y + 1)^3$.

4. Expand $(3x - 2)^3$.

5. Use the formula to evaluate $99^3$ by writing $99 = 100 - 1$.

6. Factor $x^3 + 27$ using the sum-of-cubes identity.

Answers

1. Answer to Question 1: $x^3 + 9x^2 + 27x + 27$.

2. Answer to Question 2: $a^3 - 12a^2 + 48a - 64$.

3. Answer to Question 3: $8y^3 + 12y^2 + 6y + 1$.

4. Answer to Question 4: $27x^3 - 54x^2 + 36x - 8$.

5. Answer to Question 5: $970{,}299$.

6. Answer to Question 6: $(x + 3)(x^2 - 3x + 9)$.

Key Takeaways

• The cube of a binomial expands to four terms: $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$.

• The two middle terms, each carrying a coefficient of 3, are what students drop most — and they are never zero.

• $(a-b)^3$ uses the same coefficients with alternating signs: plus, minus, plus, minus.

• The coefficients $1, 3, 3, 1$ match Pascal's triangle and the binomial theorem.

• Deriving the formula once by hand fixes both the dropped-term and missing-coefficient mistakes.

Practice these to solidify your understanding

Try these three, expanding fully and checking with a number where you can:

1. Expand $(x + 4)^3$.

2. Expand $(2m - 1)^3$.

3. Use the formula to evaluate $99^3$ by writing $99 = 100 - 1$.

If you get stuck on a middle term, return to the derivation section and rebuild the expansion line by line. At Bhanzu, our trainers introduce this formula through the physical cube model, so students see the eight blocks before they meet the four terms. Want a live trainer to work through more cube-of-a-binomial problems? Book a free demo class.