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