Cube of a Binomial

Cube of a Binomial | Formulas for Sum and Difference of Cubes | How to Expand a Cubic Binomial?

A cube of a binomial can be defined by multiplying itself three times. You can also check both the sum of cubes and the difference of cubes formulas in this article. Find out different problems on a cube of a binomial, procedure to find a cube of a binomial along with detailed steps. Learn How to Expand a Cubic Binomial in Algebraic Expressions and solve the related problems easily. Check out the solved examples on How to Cube Binomials and get to know the concept involved behind them.

Sum of Cubes Formula

The sum of a cubes of two binomials is represented as (a + b)³ = a³ + 3a²b + 3ab² + b³. Add the cube of the first term, three times the square of the first term by the second term, three times the first term by the square of the second term, and also the cube of the second term. By adding these terms you can get the Sum of cubes.

(a + b)³ = a³ + 3a²b + 3ab² + b³
= a³ + 3ab (a + b) + b³

Difference of Cubes Formula

The difference of cubes of two binomials is represented as (a – b)³ = a³ – 3a²b + 3ab² – b³. Subtract the cube of the first term and three times the square of the first term by the second term. Then, add three times the first term by the square of the second term, then subtract the cube of the second term from it.

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

Solved Problems on Expansion of Cubic Binomial

Simplify the following by cubing
1. (a + 5b)3 + (a – 5b)3

Solution:
Given expression is (a + 5b)3 + (a – 5b)3
In the given expression, the first term is in the form of (a + b) and the second term is in the form of (a – b)
Compare the first term (a + 5b)3 with (a + b) and expand it.
We know that (a + b)3 = a3 + 3a2b + 3ab2 + b3 where a = a, b = 5b
(a + 5b)3 = a3 + 3 . a2 . 5b + 3 . a . (5b)2 + (5b)3
Compare the second term (a – 5b)3 with (a – b) and expand it.
We know that (a – b)3 = a3 – 3a2b + 3ab2 – b3 where a = a, b = 5b
(a – 5b)3 = a3 – 3 . a2 . 5b + 3 . a . (5b)2 – (5b)3
Now, write (a + 5b)3 + (a – 5b)3 = a3 + 3 . a2 . 5b + 3 . a . (5b)2 + (5b)3 + a3 – 3 . a2 . 5b + 3 . a . (5b)2 – (5b)3
= a3 + 15a2b + 75ab2 + 125 b3 + a3 – 15a2b + 75ab2 – 125 b3
= 2a3 + 150ab2
The final answer is 2a3 + 150ab2

Therefore, (a + 5b)3 + (a – 5b)3 = 2a3 + 150ab2

2. (2x + 3y)3 + (2x – 3y)3

Solution:
Given expression is (2x + 3y)3 + (2x – 3y)3
In the given expression, the first term is in the form of (a + b) and the second term is in the form of (a – b)
Compare the first term (2x + 3y)3 with (a + b) and expand it.
We know that (a + b)3 = a3 + 3a2b + 3ab2 + b3 where a = 2x, b = 3y
(2x + 3y)3 = (2x)3 + 3 . (2x)2 . (3y) + 3 . (2x) . (3y)2 + (3y)3
Compare the second term (2x – 3y)3 with (a – b) and expand it.
We know that (a – b)3 = a3 – 3a2b + 3ab2 – b3 where a = 2x, b = 3y
(2x – 3y)3 = (2x)3 – 3 . (2x)2 . (3y) + 3 . (2x) . (3y)2 – (3y)3
Now, write (2x + 3y)3 + (2x – 3y)3 = (2x)3 + 3 . (2x)2 . (3y) + 3 . (2x) . (3y)2 + (3y)3 + (2x)3 – 3 . (2x)2 . (3y) + 3 . (2x) . (3y)2 – (3y)3
= 8x3 + 36x2y + 54xy2 + 27y3 + 8x3 – 36x2y + 54xy2 – 27y3
= 16x3 + 108xy2
The final answer is 16x3 + 108xy2

Therefore, (2x + 3y)3 + (2x – 3y)3 = 16x3 + 108xy2

3. (2 – 3a)3 – (5 + 3a)3

Solution:
Given expression is (2 – 3a)3 – (5 + 3a)3
In the given expression, the first term is in the form of (a – b)3 and the second term is in the form of (a + b)3
Compare the first term (2 – 3a)3 with (a – b) and expand it.
We know that (a – b)3 = a3 – 3a2b + 3ab2 – b3 where a = 2, b = 3a
(2 – 3a)3 = (2)3 – 3 . (2)2 . (3a) + 3 . (2) . (3a)2 – (3a)3
Compare the second term (5 + 3a)3 with (a + b) and expand it.
We know that (a + b)3 = a3 + 3a2b + 3ab2 + b3 where a = 5, b = 3a
(5 + 3a)3 = (5)3 + 3 . (5)2 . (3a) + 3 . (5) . (3a)2 + (3a)3
Now, write (2 – 3a)3 – (5 + 3a)3 = (2)3 – 3 . (2)2 . (3a) + 3 . (2) . (3a)2 – (3a)3 + (5)3 + 3 . (5)2 . (3a) + 3 . (5) . (3a)2 + (3a)3
= {8 – 36a + 54 a2 – 27 a3} – {125 + 225a + 135a2 + 27 a3}
=  8 – 36a + 54 a2 – 27 a3 – 125 – 225a – 135a2 – 27 a3
= 8 – 125 – 36a – 225a + 54 a2 – 135a2 – 27 a3 – 27 a3
= -117 – 261a – 81 a2 – 54 a3
The final answer is -117 – 261a – 81 a2 – 54 a3

Therefore, (2 – 3a)3 – (5 + 3a)3 = -117 – 261a – 81 a2 – 54 a3

4. (5x + 2y)3 – (5x – 2y)3

Solution:
Given expression is (5x + 2y)3 – (5x – 2y)3
In the given expression, the first term is in the form of (a + b) and the second term is in the form of (a – b)
Compare the first term (5x + 2y)3 with (a + b) and expand it.
We know that (a + b)3 = a3 + 3a2b + 3ab2 + b3 where a = 5x, b = 2y
(5x + 2y)3 = (5x)3 + 3 . (5x)2 . (2y) + 3 . (5x) . (2y)2 + (2y)3
Compare the second term (5x – 2y)3 with (a – b) and expand it.
We know that (a – b)3 = a3 – 3a2b + 3ab2 – b3 where a = 2x, b = 3y
(5x – 2y)3 = (5x)3 – 3 . (5x)2 . (2y) + 3 . (5x) . (2y)2 – (2y)3
Now, write (5x + 2y)3 – (5x – 2y)3 = (5x)3 + 3 . (5x)2 . (2y) + 3 . (5x) . (2y)2 + (2y)3 + (5x)3 – 3 . (5x)2 . (2y) + 3 . (5x) . (2y)2 – (2y)3
= {125 x3 + 150 x2 y + 60 xy2 + 8 y3} – {125 x3 – 150 x2 y + 60 xy2 – 8 y3}
= 125 x3 + 150 x2 y + 60 xy2 + 8 y3 – 125 x3 + 150 x2 y – 60 xy2 + 8 y3
= 125 x3 – 125 x3 + 150 x2 y + 150 x2 y + 60 xy2 – 60 xy2 + 8 y3 + 8 y3
= 300 x2 y + 16 y3
The final answer is 300 x2 y + 16 y3

Therefore, (5x + 2y)3 – (5x – 2y)3 = 300 x2 y + 16 y3

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