The formula (a + b)³ and (a – b)³ is used to find the cube of the binomial. The binomial theorem is an expansion of the algebraic expression. Expansion of the binomial expansion is the product of each constant term with the expression. For your convenience, we have provided the derivation of (a + b)³ and (a – b)³ along with the corollaries. Also, you can find suitable examples of Expansion of Powers of Binomials and Trinomials from this page.

## Expansion of (a ± b)^3 Derivation

**Expansion of (a + b)^3:**

(a + b)³ can be written as (a + b)(a + b)²

(a + b)³ = (a + b)(a + b)²

(a + b)³ = (a + b)[a² + 2ab + b²]

(a + b)³ = a [a² + 2ab + b²] + b [a² + 2ab + b²]

(a + b)³ = a³ + 2a²b + ab² + ba² + 2ab² + b³

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a + b)³ = a³ + 3ab(a + b) + b³

Thus the formula of (a + b)³ = a³ + 3ab(a + b) + b³

It can be read as cubed a + 3 × product of two terms (sum of the two constant terms) + cubed b

**Expansion of (a – b)^3:**

(a – b)³ can be written as (a – b)(a – b)²

(a – b)³ = (a – b)(a – b)²

(a – b)³ = (a – b) [a² + b² – 2ab]

(a – b)³ = a[a² + b² – 2ab] -b [a² + b² – 2ab]

(a – b)³ = a³ + ab² – 2a²b – ba² – b³ + 2ab²

(a – b)³ = a³ + 3ab² – 3a²b – b³

(a – b)³ = a³ + 3ab(a – b) – b³

Thus the formula of (a – b)³ = a³ + 3ab(a – b) – b³

It can be read as cubed a + 3 × product of two terms (difference of the two constant terms) – cubed b

**Corollaries:**

- (a + b)³ = a³ + 3ab(a + b) + b³
- (a – b)³ = a³ + 3ab(a – b) – b³
- (a + b)³ – (a³ + b³) = 3ab(a + b)
- (a – b)³ – (a³ – b³) = 3ab(a – b)
- a³ + b³ = (a + b)³ – 3ab(a + b)
- a³ – b³ = (a – b)³ + 3ab(a – b)

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### Expansion of (a ± b)^3 Examples

**Example 1.**

Solve the equation (x – 2y)³ using (a – b)³ formula

**Solution:**

Given that

(x – 2y)³

We know that

(a – b)³ = a³ – 3a²b + 3ab² – b³

Here

a = 1, b= -2y

Substitute a, b in the above equation then we get

x³ – 3(x)² (-2y) + 3(x)(-2y)² – (-2y)³

x³ + 6x²y – 12xy² + 8y³

Therefore the solution of the expression (x – 2y)³ is x³ + 6x²y – 12xy² + 8y³

**Example 2.**

Solve the equation (2x + 4y)³ using (a – b)³ formula

**Solution:**

Given that

(2x + 4y)³

We know that

(a – b)³ = a³ – 3a²b + 3ab² – b³

Here

a = 2x, b= 4y

Substitute a, b in the above equation then we get

2x³ – 3(2x)² (4y) + 3(2x)(4y)² – (4y)³

2x³ – 24x²y + 24xy² – 64y³

Therefore the solution of the expression (2x + 4y)³ is 2x³ – 24x²y + 24xy² – 64y³

**Example 3.**

Solve the equation (2x + y)³ using the formula (a + b)³

**Solution:**

Given that

(2x + y)³

We know that

(a + b)³ = a³ + 3ab(a + b) + b²

Here a = 2x, b = y

Substitute a, b in the above equation then we get

(2x)³ + 3(2x)(y)(2x + y) + y³

2x³ + 6xy(2x + y) + y³

2x³ + 12x²y + 6xy² + y³

Therefore the solution of the expression (2x + y)³ is 2x³ + 12x²y + 6xy² + y³

**Example 4.**

Solve the equation (3x + 2y)³ using the formula (a + b)³

**Solution:**

Given that

(3x + 2y)³

We know that

(a + b)³ = a³ + 3a²b + 3ab² + b³

Here

a= 3x, b= 2y

Substitute a, b in the above equation then we get

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

27x³ + 3(9x²)(2y) + 3(3x)(4y²) + 8y³

27x³ + 54x²y + 36xy² + 8y³

Therefore the solution of the expression (2x + y)³ is 27x³ + 54x²y + 36xy² + 8y³

**Example 5.**

Solve the equation (4x + 2y)³ using the formula (a + b)³

**Solution:**

Given that

(4x + 2y)³

We know that

(a + b)³ = a³ + 3ab(a + b) + b²

Here a = 4x, b = 2y

Substitute a, b in the above equation then we get

(4x)³ + 3(4x)(2y)(4x + 2y) + 2y³

4x³ + 24xy(4x + 2y) + 2y³

4x³ + 96x²y + 48xy² + 2y³

Therefore the solution of the expression (4x + 2y)³ is 4x³ + 96x²y + 48xy² + 2y³