(a + b)³ Formula: A Plus B Whole Cube Formula, Proof, Examples & Exercises

(a + b)³ Formula: A Plus B Whole Cube Formula, Proof, Examples & Exercises

Complete Guide to the (a + b) Whole Cube Formula – Derivation, Algebraic Identities, (a − b)³, a³ + b³, a³ − b³, (a + b + c)³, Solved Examples, Practice Questions and Answers

Victoria Academy – A Canadian International School
Ontario, Canada | Grades 9–12 | BSID #884963

The (a + b)³ formula, also called the a plus b whole cube formula, is one of the most useful algebraic identities students learn.

The formula is:

(a + b)³ = a³ + 3a²b + 3ab² + b³

It can also be written as:

(a + b)³ = a³ + b³ + 3ab(a + b)

This formula helps students:

expand algebraic expressions;

simplify polynomial calculations;

solve algebra problems;

factor expressions;

understand binomial expansion;

calculate cubes mentally;

connect algebra with geometry;

and prepare for more advanced mathematics.

The most important thing is not simply to memorize the formula.

Students should understand:

WHY THE FORMULA WORKS.

This guide explains the formula step by step, includes related cube identities, solved examples, common mistakes and practice exercises with answers. (Sunbeam World School)

Quick Answer: What Is the (a + b)³ Formula?

The formula is:

(a + b)³ = a³ + 3a²b + 3ab² + b³

Another useful form is:

(a + b)³ = a³ + b³ + 3ab(a + b)

The coefficients follow the pattern:

1 : 3 : 3 : 1

So we can remember:

a³ + 3a²b + 3ab² + b³

What Does (a + b)³ Mean?

The expression:

(a + b)³

means:

(a + b)(a + b)(a + b)

It does not mean:

a³ + b³

This is one of the most common mistakes students make.

For example:

(2 + 3)³

= 5³

= 125

But:

2³ + 3³

= 8 + 27

= 35

Clearly:

125 ≠ 35

Therefore:

(a + b)³ ≠ a³ + b³

The middle terms are essential.

What Is an Algebraic Identity?

An algebraic identity is an equality that is true for all permissible values of the variables involved.

For example:

(a + b)² = a² + 2ab + b²

and:

(a + b)³ = a³ + 3a²b + 3ab² + b³

are identities.

This means the relationship remains true no matter what suitable values we substitute for a and b.

Derivation of the (a + b)³ Formula

Students do not need to memorize the formula blindly.

We can derive it.

Start with:

(a + b)³

By definition:

(a + b)³ = (a + b)(a + b)(a + b)

We already know:

(a + b)² = a² + 2ab + b²

Therefore:

(a + b)³

= (a + b)²(a + b)

= (a² + 2ab + b²)(a + b)

Now multiply each term.

First multiply by a:

a(a² + 2ab + b²)

= a³ + 2a²b + ab²

Now multiply by b:

b(a² + 2ab + b²)

= a²b + 2ab² + b³

Add the results:

a³ + 2a²b + ab² + a²b + 2ab² + b³

Combine like terms:

2a²b + a²b = 3a²b

and:

ab² + 2ab² = 3ab²

Therefore:

(a + b)³ = a³ + 3a²b + 3ab² + b³

That is the a plus b whole cube formula. (Sunbeam World School)

Short Derivation of (a + b)³

For quick revision:

(a + b)³

= (a + b)²(a + b)

= (a² + 2ab + b²)(a + b)

= a³ + a²b + 2a²b + 2ab² + ab² + b³

= a³ + 3a²b + 3ab² + b³

So:

(a + b)³ = a³ + 3a²b + 3ab² + b³

Compact Form of the (a + b)³ Formula

Start with:

(a + b)³

= a³ + 3a²b + 3ab² + b³

Factor the middle two terms:

3a²b + 3ab²

= 3ab(a + b)

Therefore:

(a + b)³ = a³ + b³ + 3ab(a + b)

This form is especially useful when we know:

a + b

and:

ab

but do not know the individual values of a and b.

How to Remember the (a + b)³ Formula

A useful pattern is:

1 – 3 – 3 – 1

Then look at the powers.

For a:

a³ → a² → a¹ → a⁰

The power decreases.

For b:

b⁰ → b¹ → b² → b³

The power increases.

So:

a³ + 3a²b + 3ab² + b³

Another pattern:

First term:

a³

Second:

3 × a² × b

Third:

3 × a × b²

Fourth:

b³

Pascal's Triangle and the (a + b)³ Formula

The coefficients:

1, 3, 3, 1

come from the corresponding row of Pascal's Triangle.

A simplified Pascal's Triangle begins:

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

The pattern:

1 3 3 1

gives the coefficients of:

(a + b)³

Therefore:

(a + b)³

= 1a³ + 3a²b + 3ab² + 1b³

= a³ + 3a²b + 3ab² + b³. (Sunbeam World School)

What Is the (a − b)³ Formula?

Another important identity is:

(a − b)³ = a³ − 3a²b + 3ab² − b³

Notice the signs:

+ − + −

A useful compact form is:

(a − b)³ = a³ − b³ − 3ab(a − b)

Derivation of (a − b)³

Start with:

(a − b)³

= (a − b)²(a − b)

We know:

(a − b)²

= a² − 2ab + b²

Therefore:

(a − b)³

= (a² − 2ab + b²)(a − b)

Multiply:

= a³ − a²b − 2a²b + 2ab² + ab² − b³

Combine like terms:

= a³ − 3a²b + 3ab² − b³

Therefore:

(a − b)³ = a³ − 3a²b + 3ab² − b³

(a + b)³ vs (a − b)³

Compare them carefully:

A Plus B Whole Cube

(a + b)³ = a³ + 3a²b + 3ab² + b³

A Minus B Whole Cube

(a − b)³ = a³ − 3a²b + 3ab² − b³

Notice:

For the plus formula:

+ + + +

For the minus formula:

+ − + −

What Is the Formula for a³ + b³?

The sum of cubes formula is:

a³ + b³ = (a + b)(a² − ab + b²)

This is different from:

(a + b)³

Students frequently confuse these formulas.

Remember:

Whole Cube

(a + b)³ = a³ + 3a²b + 3ab² + b³

Sum of Cubes

a³ + b³ = (a + b)(a² − ab + b²)

They are related, but they are not the same identity. (Sunbeam World School)

What Is the Formula for a³ − b³?

The difference of cubes formula is:

a³ − b³ = (a − b)(a² + ab + b²)

Compare:

Sum of Cubes

a³ + b³ = (a + b)(a² − ab + b²)

Difference of Cubes

a³ − b³ = (a − b)(a² + ab + b²)

Easy Way to Remember Sum and Difference of Cubes

For:

a³ + b³

use:

(a + b)

then:

a² − ab + b²

So:

a³ + b³ = (a + b)(a² − ab + b²)

For:

a³ − b³

use:

(a − b)

then:

a² + ab + b²

So:

a³ − b³ = (a − b)(a² + ab + b²)

Notice the sign inside the quadratic factor is opposite to the original sign.

How Are (a + b)³ and a³ + b³ Connected?

From:

(a + b)³

= a³ + b³ + 3ab(a + b)

we can rearrange:

a³ + b³ = (a + b)³ − 3ab(a + b)

This is extremely useful when:

a + b

and:

ab

are known.

How Are (a − b)³ and a³ − b³ Connected?

We know:

(a − b)³

= a³ − b³ − 3ab(a − b)

Therefore:

a³ − b³ = (a − b)³ + 3ab(a − b)

What Is the (a + b + c)³ Formula?

The cube of three terms can be written as:

(a + b + c)³ = a³ + b³ + c³ + 3(a + b)(b + c)(c + a)

The fully expanded version is:

(a + b + c)³

= a³ + b³ + c³

+ 3a²b + 3a²c

+ 3ab² + 3b²c

+ 3ac² + 3bc²

+ 6abc

This identity extends the idea of cubing a binomial to three terms. (Sunbeam World School)

Another Important Identity With a³ + b³ + c³

A useful identity is:

a³ + b³ + c³ − 3abc

= (a + b + c)(a² + b² + c² − ab − bc − ca)

This has an important special case.

If:

a + b + c = 0

then:

a³ + b³ + c³ = 3abc

All Important Cube Formulas in One Table

Identity

Formula

(a + b)³

a³ + 3a²b + 3ab² + b³

Compact (a + b)³

a³ + b³ + 3ab(a + b)

(a − b)³

a³ − 3a²b + 3ab² − b³

Compact (a − b)³

a³ − b³ − 3ab(a − b)

a³ + b³

(a + b)(a² − ab + b²)

a³ − b³

(a − b)(a² + ab + b²)

(a + b + c)³

a³ + b³ + c³ + 3(a + b)(b + c)(c + a)

a³ + b³ + c³ − 3abc

(a + b + c)(a² + b² + c² − ab − bc − ca)

Solved Example 1 – Expand (x + 2)³

Use:

(a + b)³

= a³ + 3a²b + 3ab² + b³

Here:

a = x

b = 2

Therefore:

(x + 2)³

= x³ + 3(x²)(2) + 3(x)(2²) + 2³

= x³ + 6x² + 12x + 8

Answer:

(x + 2)³ = x³ + 6x² + 12x + 8

Solved Example 2 – Expand (x + 5)³

Here:

a = x

b = 5

Apply the formula:

(x + 5)³

= x³ + 3x²(5) + 3x(25) + 125

= x³ + 15x² + 75x + 125

Answer:

x³ + 15x² + 75x + 125

Solved Example 3 – Expand (2x + 3)³

Here:

a = 2x

b = 3

Use:

(a + b)³ = a³ + 3a²b + 3ab² + b³

Therefore:

(2x + 3)³

= (2x)³ + 3(2x)²(3) + 3(2x)(3²) + 3³

= 8x³ + 36x² + 54x + 27

Answer:

8x³ + 36x² + 54x + 27

Solved Example 4 – Expand (2x + 3y)³

Let:

a = 2x

b = 3y

Then:

(2x + 3y)³

= (2x)³ + 3(2x)²(3y) + 3(2x)(3y)² + (3y)³

= 8x³ + 36x²y + 54xy² + 27y³

Answer:

8x³ + 36x²y + 54xy² + 27y³

Solved Example 5 – Expand (x − 4)³

Use:

(a − b)³

= a³ − 3a²b + 3ab² − b³

Here:

a = x

b = 4

Therefore:

(x − 4)³

= x³ − 3x²(4) + 3x(16) − 64

= x³ − 12x² + 48x − 64

Answer:

x³ − 12x² + 48x − 64

Solved Example 6 – Expand (3x − 2)³

Use:

a = 3x

b = 2

Then:

(3x − 2)³

= (3x)³ − 3(3x)²(2) + 3(3x)(2²) − 2³

= 27x³ − 54x² + 36x − 8

Answer:

27x³ − 54x² + 36x − 8

Solved Example 7 – Find 11³ Using the Formula

Instead of multiplying:

11 × 11 × 11,

write:

11 = 10 + 1

Therefore:

11³

= (10 + 1)³

= 10³ + 3(10²)(1) + 3(10)(1²) + 1³

= 1000 + 300 + 30 + 1

= 1331

Answer:

11³ = 1331

Solved Example 8 – Find 21³ Mentally

Write:

21 = 20 + 1

Then:

21³

= (20 + 1)³

= 20³ + 3(20²)(1) + 3(20)(1²) + 1

= 8000 + 1200 + 60 + 1

= 9261

Answer:

21³ = 9261

Solved Example 9 – Find 99³

We could use:

99 = 100 − 1

Use:

(a − b)³

Therefore:

99³

= (100 − 1)³

= 100³ − 3(100²)(1) + 3(100)(1²) − 1

= 1,000,000 − 30,000 + 300 − 1

= 970,299

Answer:

99³ = 970,299

Solved Example 10 – If a + b = 5 and ab = 6, Find a³ + b³

Use:

a³ + b³ = (a + b)³ − 3ab(a + b)

Substitute:

a + b = 5

ab = 6

Therefore:

a³ + b³

= 5³ − 3(6)(5)

= 125 − 90

= 35

Answer:

a³ + b³ = 35

Solved Example 11 – If a + b = 10 and ab = 21, Find a³ + b³

Use:

a³ + b³

= (a + b)³ − 3ab(a + b)

= 10³ − 3(21)(10)

= 1000 − 630

= 370

Answer:

a³ + b³ = 370

Solved Example 12 – If a − b = 4 and ab = 5, Find a³ − b³

Use:

a³ − b³ = (a − b)³ + 3ab(a − b)

Substitute:

= 4³ + 3(5)(4)

= 64 + 60

= 124

Answer:

a³ − b³ = 124

Solved Example 13 – Factorize x³ + 8

Recognize:

8 = 2³

Therefore:

x³ + 8

= x³ + 2³

Use:

a³ + b³

= (a + b)(a² − ab + b²)

Therefore:

x³ + 8 = (x + 2)(x² − 2x + 4)

Solved Example 14 – Factorize x³ − 27

Recognize:

27 = 3³

Therefore:

x³ − 27

= x³ − 3³

Use:

a³ − b³

= (a − b)(a² + ab + b²)

Therefore:

x³ − 27 = (x − 3)(x² + 3x + 9)

Solved Example 15 – Factorize 8x³ + 27y³

Recognize:

8x³ = (2x)³

27y³ = (3y)³

Therefore:

8x³ + 27y³

= (2x)³ + (3y)³

Use:

a³ + b³

= (a + b)(a² − ab + b²)

So:

= (2x + 3y)[(2x)² − (2x)(3y) + (3y)²]

= (2x + 3y)(4x² − 6xy + 9y²)

Answer:

(2x + 3y)(4x² − 6xy + 9y²)

Solved Example 16 – Factorize 64x³ − y³

Recognize:

64x³ = (4x)³

Therefore:

64x³ − y³

= (4x)³ − y³

Using difference of cubes:

= (4x − y)[(4x)² + (4x)(y) + y²]

Answer:

(4x − y)(16x² + 4xy + y²)

Why the Coefficients Are 1, 3, 3, 1

Think about:

(a + b)(a + b)(a + b)

To obtain:

a³

we choose a from all three brackets.

There is:

1 way

To obtain:

a²b

we choose b from one bracket and a from the other two.

There are:

3 ways

Therefore:

3a²b

To obtain:

ab²

we choose a from one bracket and b from two.

Again:

3 ways

Therefore:

3ab²

To obtain:

b³

we choose b from all three brackets.

There is:

1 way

Therefore the coefficients are:

1, 3, 3, 1

Geometric Meaning of (a + b)³

The formula can also be understood geometrically.

Consider a cube whose side length is:

a + b

The volume of the large cube is:

(a + b)³

If the cube is divided according to lengths a and b, its pieces can be grouped into:

one cube of volume:

a³

three rectangular solids with volume:

a²b

each;

three rectangular solids with volume:

ab²

each;

and one small cube of volume:

b³

Therefore total volume:

a³ + 3a²b + 3ab² + b³

Hence:

(a + b)³ = a³ + 3a²b + 3ab² + b³

This gives a visual reason for the formula rather than treating it only as something to memorize. (Sunbeam World School)

Where Is the (a + b)³ Formula Used?

The identity appears in many areas of mathematics.

Students may use it for:

Algebraic Expansion

Expanding binomials.

Factorization

Recognizing and simplifying expressions.

Mental Mathematics

Calculating cubes near convenient numbers.

Polynomial Problems

Working with cubic expressions.

Geometry

Understanding volumes.

Coordinate Mathematics

Simplifying algebraic expressions.

Calculus Preparation

Building fluency with powers and polynomial manipulation.

Higher Mathematics

Understanding the Binomial Theorem.

Connection With the Binomial Theorem

The general Binomial Theorem says:

(a + b)ⁿ = Σ C(n,r)aⁿ⁻ʳbʳ

For:

n = 3

the coefficients are:

C(3,0) = 1

C(3,1) = 3

C(3,2) = 3

C(3,3) = 1

Therefore:

(a + b)³

= a³ + 3a²b + 3ab² + b³

So the familiar cube formula is actually a special case of a much larger mathematical idea.

Common Mistakes With the (a + b)³ Formula

Mistake 1 – Writing (a + b)³ = a³ + b³

Incorrect:

(a + b)³ = a³ + b³

Correct:

(a + b)³ = a³ + 3a²b + 3ab² + b³

Never forget the middle terms.

Mistake 2 – Forgetting the Coefficient 3

Incorrect:

a³ + a²b + ab² + b³

Correct:

a³ + 3a²b + 3ab² + b³

Mistake 3 – Mixing Up the Powers

Incorrect:

a³ + 3ab + 3ab + b³

Correct:

a³ + 3a²b + 3ab² + b³

The total degree of every term is 3.

Check:

a³ → degree 3

a²b → 2 + 1 = 3

ab² → 1 + 2 = 3

b³ → degree 3

Mistake 4 – Wrong Signs in (a − b)³

Incorrect:

a³ − 3a²b − 3ab² − b³

Correct:

a³ − 3a²b + 3ab² − b³

Remember:

+ − + −

Mistake 5 – Cubing Only the Variable

For:

(2x + 3)³

a student may incorrectly start with:

2x³ + ...

But:

(2x)³ = 8x³

The entire term must be cubed.

Mistake 6 – Squaring Only the Variable in the Middle Term

For:

3(2x)²(3)

remember:

(2x)² = 4x²

not:

2x².

Mistake 7 – Confusing Whole Cube and Sum of Cubes

These are different:

(a + b)³

and:

a³ + b³

Whole cube means expansion.

Sum of cubes usually means factorization.

Quick Accuracy Check

After expanding:

(a + b)³,

check:

1. Are there four terms?

2. Are the coefficients 1, 3, 3, 1?

3. Does the power of a decrease?

3 → 2 → 1 → 0

4. Does the power of b increase?

0 → 1 → 2 → 3

5. Is every term total degree 3?

If yes, your expansion is probably correct.

Exercise 1 – Expand Using (a + b)³

Try these before checking the answers.

1. (x + 1)³

2. (x + 3)³

3. (x + 5)³

4. (2x + 1)³

5. (3x + 2)³

6. (x + 2y)³

7. (2x + y)³

8. (2a + 3b)³

9. (3m + 4n)³

10. (5x + 2y)³

Answers – Exercise 1

1. (x + 1)³

x³ + 3x² + 3x + 1

2. (x + 3)³

x³ + 9x² + 27x + 27

3. (x + 5)³

x³ + 15x² + 75x + 125

4. (2x + 1)³

8x³ + 12x² + 6x + 1

5. (3x + 2)³

27x³ + 54x² + 36x + 8

6. (x + 2y)³

x³ + 6x²y + 12xy² + 8y³

7. (2x + y)³

8x³ + 12x²y + 6xy² + y³

8. (2a + 3b)³

8a³ + 36a²b + 54ab² + 27b³

9. (3m + 4n)³

27m³ + 108m²n + 144mn² + 64n³

10. (5x + 2y)³

125x³ + 150x²y + 60xy² + 8y³

Exercise 2 – Expand Using (a − b)³

1. (x − 1)³

2. (x − 2)³

3. (x − 5)³

4. (2x − 1)³

5. (3x − 2)³

6. (x − 2y)³

7. (2a − 3b)³

8. (4m − n)³

Answers – Exercise 2

1.

x³ − 3x² + 3x − 1

2.

x³ − 6x² + 12x − 8

3.

x³ − 15x² + 75x − 125

4.

8x³ − 12x² + 6x − 1

5.

27x³ − 54x² + 36x − 8

6.

x³ − 6x²y + 12xy² − 8y³

7.

8a³ − 36a²b + 54ab² − 27b³

8.

64m³ − 48m²n + 12mn² − n³

Exercise 3 – Mental Mathematics With Cube Formulas

Use algebraic identities.

1. 9³

2. 11³

3. 19³

4. 21³

5. 29³

6. 31³

7. 49³

8. 51³

9. 99³

10. 101³

Answers – Exercise 3

1. 9³ = 729

2. 11³ = 1331

3. 19³ = 6859

4. 21³ = 9261

5. 29³ = 24,389

6. 31³ = 29,791

7. 49³ = 117,649

8. 51³ = 132,651

9. 99³ = 970,299

10. 101³ = 1,030,301

Exercise 4 – Sum and Difference of Cubes

Factorize:

1. x³ + 1

2. x³ − 1

3. x³ + 8

4. x³ − 8

5. x³ + 27

6. x³ − 27

7. 8x³ + y³

8. 8x³ − y³

9. 27a³ + 8b³

10. 64x³ − 125y³

Answers – Exercise 4

1.

x³ + 1

= x³ + 1³

= (x + 1)(x² − x + 1)

2.

= (x − 1)(x² + x + 1)

3.

= (x + 2)(x² − 2x + 4)

4.

= (x − 2)(x² + 2x + 4)

5.

= (x + 3)(x² − 3x + 9)

6.

= (x − 3)(x² + 3x + 9)

7.

= (2x + y)(4x² − 2xy + y²)

8.

= (2x − y)(4x² + 2xy + y²)

9.

= (3a + 2b)(9a² − 6ab + 4b²)

10.

= (4x − 5y)(16x² + 20xy + 25y²)

Exercise 5 – Find a³ + b³

Use:

a³ + b³ = (a + b)³ − 3ab(a + b)

1. a + b = 4, ab = 3

2. a + b = 6, ab = 5

3. a + b = 8, ab = 12

4. a + b = 10, ab = 16

5. a + b = 12, ab = 20

Answers – Exercise 5

1.

4³ − 3(3)(4)

= 64 − 36

= 28

2.

6³ − 3(5)(6)

= 216 − 90

= 126

3.

8³ − 3(12)(8)

= 512 − 288

= 224

4.

10³ − 3(16)(10)

= 1000 − 480

= 520

5.

12³ − 3(20)(12)

= 1728 − 720

= 1008

Exercise 6 – Find a³ − b³

Use:

a³ − b³ = (a − b)³ + 3ab(a − b)

1. a − b = 2, ab = 3

2. a − b = 3, ab = 4

3. a − b = 4, ab = 5

4. a − b = 5, ab = 6

5. a − b = 6, ab = 8

Answers – Exercise 6

1.

2³ + 3(3)(2)

= 8 + 18

= 26

2.

3³ + 3(4)(3)

= 27 + 36

= 63

3.

4³ + 3(5)(4)

= 64 + 60

= 124

4.

5³ + 3(6)(5)

= 125 + 90

= 215

5.

6³ + 3(8)(6)

= 216 + 144

= 360

Exercise 7 – Identify the Correct Formula

Choose the correct identity.

1. (a + b)³

A. a³ + b³
B. a³ + 3a²b + 3ab² + b³
C. a³ + 2ab + b³

Answer: B

2. a³ + b³

A. (a + b)(a² − ab + b²)
B. (a − b)(a² + ab + b²)
C. (a + b)³

Answer: A

3. a³ − b³

A. (a + b)(a² − ab + b²)
B. (a − b)(a² + ab + b²)
C. (a − b)³

Answer: B

4. (a − b)³

A. a³ − 3a²b + 3ab² − b³
B. a³ − b³
C. a³ + 3a²b − 3ab² − b³

Answer: A

Challenge Questions

Question 1

If:

a + b = 7

and:

ab = 10

find:

a³ + b³.

Solution

= (a + b)³ − 3ab(a + b)

= 7³ − 3(10)(7)

= 343 − 210

= 133

Question 2

If:

x + 1/x = 3

find:

x³ + 1/x³.

Use:

a³ + b³

= (a + b)³ − 3ab(a + b)

Here:

a = x

b = 1/x

Therefore:

ab = 1

and:

a + b = 3

So:

x³ + 1/x³

= 3³ − 3(1)(3)

= 27 − 9

= 18

Question 3

If:

x − 1/x = 2

find:

x³ − 1/x³.

Here:

ab = x(1/x) = 1

Use:

a³ − b³

= (a − b)³ + 3ab(a − b)

= 2³ + 3(1)(2)

= 8 + 6

= 14

7-Day Cube Formula Practice Plan

Day 1 – Memorize With Understanding

Learn:

(a + b)³

and derive it once.

Day 2 – Practise (a + b)³

Complete 10 expansions.

Day 3 – Practise (a − b)³

Focus on the sign pattern:

+ − + −

Day 4 – Sum and Difference of Cubes

Learn:

a³ + b³

and:

a³ − b³.

Day 5 – Mental Mathematics

Find cubes of numbers near:

10

20

50

Day 6 – Word and Algebra Problems

Practise questions where:

a + b

and:

ab

are given.

Day 7 – Mixed Revision

Mix:

expansion;

factorization;

mental arithmetic;

identities;

and algebraic applications.

Best Way to Learn Algebraic Identities

Do not memorize ten formulas as unrelated information.

Look for patterns.

For the cube formula:

1 – 3 – 3 – 1

For the powers:

a³ → a² → a → 1

while:

1 → b → b² → b³

For (a − b)³:

+ − + −

Understanding patterns reduces memory load and makes mistakes easier to detect.

Why Algebraic Identities Matter

Algebraic identities help students develop more than formula memorization.

They strengthen:

Pattern Recognition

Logical Reasoning

Algebraic Manipulation

Problem-Solving

Mental Mathematics

Mathematical Communication

Preparation for Advanced Mathematics

A student who understands why an identity works is better prepared to apply it in unfamiliar problems.

Mathematics Learning at Victoria Academy

Victoria Academy

A Canadian International School

Ontario, Canada | Grades 9–12 | BSID #884963

At Victoria Academy, mathematics learning can involve more than memorizing procedures.

Students studying Ontario high-school mathematics are encouraged to develop:

Mathematical Reasoning

Problem-Solving

Communication

Critical Thinking

Application of Knowledge

Algebraic Fluency

Data and Mathematical Analysis

Independent Learning

Strong mathematics learning should help students answer two questions:

“WHAT IS THE FORMULA?”

and:

“WHY DOES THE FORMULA WORK?”

Understanding both can help students apply mathematical ideas more confidently in new situations.

Algebra and Future STEM Pathways

Strong algebraic foundations can become particularly important for students considering future studies in areas such as:

Engineering

Computer Science

Artificial Intelligence

Data Science

Mathematics

Physics

Economics

Statistics

Finance

Business Analytics

Architecture-related programs

and other STEM-oriented pathways.

Different universities and programs have different senior-secondary mathematics prerequisites.

Students should therefore plan Grade 11 and Grade 12 mathematics courses with their future university goals in mind.

Frequently Asked Questions About the (a + b)³ Formula

1. What is the formula for (a + b)³?

(a + b)³ = a³ + 3a²b + 3ab² + b³

2. What is another form of (a + b)³?

(a + b)³ = a³ + b³ + 3ab(a + b)

3. What is the (a − b)³ formula?

(a − b)³ = a³ − 3a²b + 3ab² − b³

4. What is the compact form of (a − b)³?

(a − b)³ = a³ − b³ − 3ab(a − b)

5. Is (a + b)³ equal to a³ + b³?

No.

The correct formula is:

a³ + 3a²b + 3ab² + b³

6. Why does (a + b)³ have four terms?

Because expanding:

(a + b)(a + b)(a + b)

and combining like terms produces four distinct terms:

a³

a²b

ab²

b³.

7. What are the coefficients in (a + b)³?

1, 3, 3, 1

8. Where do 1, 3, 3, 1 come from?

They appear in Pascal's Triangle and also arise from the number of ways terms can be selected from the three binomial factors.

9. What is the sum of cubes formula?

a³ + b³ = (a + b)(a² − ab + b²)

10. What is the difference of cubes formula?

a³ − b³ = (a − b)(a² + ab + b²)

11. What is the difference between (a + b)³ and a³ + b³?

(a + b)³ is the cube of a sum.

a³ + b³ is the sum of two separate cubes.

They are different expressions.

12. How do I expand (x + 2)³?

x³ + 6x² + 12x + 8

13. How do I expand (x − 2)³?

x³ − 6x² + 12x − 8

14. What is (2x + y)³?

8x³ + 12x²y + 6xy² + y³

15. What is (2x − y)³?

8x³ − 12x²y + 6xy² − y³

16. What is (a + b + c)³?

A useful form is:

a³ + b³ + c³ + 3(a + b)(b + c)(c + a)

17. What happens if a + b + c = 0?

Using:

a³ + b³ + c³ − 3abc

= (a + b + c)(...)

if:

a + b + c = 0

then:

a³ + b³ + c³ = 3abc

18. Can the (a + b)³ formula be used for mental maths?

Yes.

For example:

21³

can be written as:

(20 + 1)³.

19. How can I calculate 99³ quickly?

Use:

99 = 100 − 1

and apply:

(a − b)³.

The answer is:

970,299

20. How do I find a³ + b³ if a + b and ab are known?

Use:

a³ + b³ = (a + b)³ − 3ab(a + b)

21. How do I find a³ − b³ if a − b and ab are known?

Use:

a³ − b³ = (a − b)³ + 3ab(a − b)

22. What is the easiest way to remember the plus-cube formula?

Remember:

1 – 3 – 3 – 1

and powers:

a³, a²b, ab², b³.

23. How do I remember the minus-cube formula?

Use the same coefficients but alternate signs:

+ − + −

24. Is the (a + b)³ formula part of the Binomial Theorem?

Yes.

It is the case:

n = 3

of the Binomial Theorem.

25. Why should students understand the derivation?

Understanding the derivation helps students remember the identity, recognize patterns and apply it to unfamiliar algebra problems.

26. Does every term in the expansion have degree 3?

Yes.

For example:

a²b

has total degree:

2 + 1 = 3.

And:

ab²

has total degree:

1 + 2 = 3.

27. What is the factorization of x³ + 8?

(x + 2)(x² − 2x + 4)

28. What is the factorization of x³ − 8?

(x − 2)(x² + 2x + 4)

29. What is the factorization of x³ + 27?

(x + 3)(x² − 3x + 9)

30. What is the factorization of x³ − 27?

(x − 3)(x² + 3x + 9)

Quick Revision Sheet

A Plus B Whole Cube

(a + b)³

= a³ + 3a²b + 3ab² + b³

= a³ + b³ + 3ab(a + b)

A Minus B Whole Cube

(a − b)³

= a³ − 3a²b + 3ab² − b³

= a³ − b³ − 3ab(a − b)

Sum of Cubes

a³ + b³

= (a + b)(a² − ab + b²)

Difference of Cubes

a³ − b³

= (a − b)(a² + ab + b²)

Three-Term Identity

a³ + b³ + c³ − 3abc

= (a + b + c)(a² + b² + c² − ab − bc − ca)

If:

a + b + c = 0

then:

a³ + b³ + c³ = 3abc

Final Student Tip

Do not learn the formula as a random line of symbols.

See the pattern:

1 – 3 – 3 – 1

Watch the powers:

a³ → a²b → ab² → b³

Understand the multiplication.

Practise expanding different expressions.

Then learn how the same identity connects to:

mental mathematics;

factorization;

sum of cubes;

difference of cubes;

Pascal's Triangle;

geometry;

and the Binomial Theorem.

The goal is not only:

MEMORIZE THE FORMULA.

The stronger goal is:

UNDERSTAND THE PATTERN.

APPLY IT CORRECTLY.

CHECK YOUR WORK.

USE MATHEMATICS TO SOLVE NEW PROBLEMS.

About Victoria Academy

Victoria Academy

A Canadian International School

Ontario, Canada

Grades 9–12

BSID #884963

Victoria Academy provides Ontario high-school learning opportunities for eligible Grades 9–12 students, including international students who may be able to study online from their home country.

Students studying mathematics through an Ontario high-school pathway can build mathematical reasoning, problem-solving, communication and analytical skills that support future study in mathematics, science, technology, business and other university pathways.

Students pursuing the Ontario Secondary School Diploma (OSSD) must successfully complete all applicable Ontario graduation requirements based on their individual academic status and approved academic plan.

STAY LOCAL. STUDY CANADIAN. THINK GLOBAL.

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For internal SEO linking, this page can connect naturally with:

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4. a³ − b³ Formula: Difference of Cubes With Examples

5. (a + b)² Formula: A Plus B Whole Square

6. (a − b)² Formula: A Minus B Whole Square

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What is the formula for (a + b)³?

The formula is:

(a + b)³ = a³ + 3a²b + 3ab² + b³.

It can also be written as:

(a + b)³ = a³ + b³ + 3ab(a + b).

The coefficients follow the pattern 1, 3, 3, 1.

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Economics

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Eligible international students can explore Ontario high-school education with:

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A Canadian International School
Ontario, Canada | Grades 9–12 | BSID #884963

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