The sum and difference of cubes are algebraic formulas used to factor expressions of the form a3+b3 and a3−b3 respectively. These formulas are particularly useful in simplifying and solving polynomial equations.
It is the basic formula of algebra used to solve the sum of the cubes and the difference of the cubes without actually calculating the values of the cubes. The sum of the cubes of the polynomial is represented as, a3 + b3 whereas the difference of the cubes is represented as a3- b3. These algebraic expressions are easily factorized using various algebraic expressions without actually calculating the cubes.
In this article, we will learn about Sum of Cubes, Sum of Cubes Formula, Factoring Sum of Cubes, Difference of Cubes, Difference of Cubes Formula with examples in detail below.
Sum of Cubes
Sum of cubes is the formula that is used to find the sum of two cubes without actually finding their cubes arithmetically. The sum of cubes is very useful in solving various algebraic problems and is helpful in quickly solving various problems. The sum of cubes formula is the formula that is used to factorize the sum of cubes, The formula for the sum of cubes is discussed below:
Sum of Cubes Formula
Cube of a number is the number multiplied by itself twice. The sum of the cube is the formula that is the formula for a3 + b3 and its formula is added below,
a3 + b3 = (a + b)(a2 – ab + b2)
The above formula is algebraic identity and that is used to find the sum of cubes formula.
Sum of Cube Formula Proof
This identity can be proved by multiplying the expressions on the right side and getting equal to the left side expression. Here is the proof of this identity.
Given Identity:
a3 + b3 = (a + b) (a2 - ab + b2)
Proof:
= RHS
= (a + b)(a2 – ab + b2)
= a(a2 – ab + b2)) + b(a2 – ab + b2)
= a3 – a2b + ab2 + a2b – ab2 + b3
= a3 – a2b + a2b + ab2 – ab2 + b3
= a3 + b3
= LHS
Hence proved.
Factoring Sum of Cubes
We use the sum of cubes formula to easily factorize the cubes in polynomials. This is explained by the example added below,
For example, suppose we have to factorize, x3 + 27
Solution:
= x3 + 27
= x3 + 33
Using Identity, a3 + b3 = (a + b) (a2 - ab + b2)
where,
- a = x
- b = 3
= (x + 3)(x2 -(x)(3) + 32)
= (x + 3)(x2 - 3x + 9)
Thus, the factors of x3 + 27 are easily found.
Difference of Cubes
When subtracting any two polynomials, a3 - b3, the difference of cubes formula is utilized. This formula is easy to memorize and may be completed in minutes. It is similar to how the sum of cubes formula works.
Difference of Cube Formula
a3 – b3 = (a – b) (a2 + ab + b2)
Difference of Cube Formula Proof
This identity can be proved by multiplying the expressions on the right side and getting equal to the left side expression. Here is the proof of this identity.
Given Identity:
a3 – b3 = (a – b) (a2 + ab + b2)
Proof:
= RHS
= (a - b)(a2 + ab + b2)
= a(a2 + ab + b2)) - b(a2 + ab + b2)
= a3 + a2b + ab2 - a2b - ab2 - b3
= a3 – a2b + a2b + ab2 – ab2 - b3
= a3 - b3
= LHS
Hence proved.
Factoring Difference of Cubes
We use the difference of cubes formula to easily factorize the cubes in polynomials. This is explained by the example added below:
For example, suppose we have to factorize, x3 – 343
Solution:
= x3 - 343
= x3 – 73
Using identity a3 – b3 = (a – b) (a2 + ab + b2)
where,
- a = x
- b = 7
= (x – 7) (x2 + (x)(7) + 72)
= (x – 7) (x2 + 7x + 49)
Thus, the factors of x3 - 343 are easily found.
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Examples on Sum and Difference of Cubes
Example 1: Factorize y3 – 125
Solution:
y3 – 125 = y3 – 53
Since, a3 – b3 = (a – b) (a2 + ab + b2),
here,
- a = y
- b = 5
= (y – 5) (y2 + (y)(5) + 52)
= (y – 5) (y2 + 5y + 25)
Example 2: Evaluate 253 – 123
Solution:
Since, a3 – b3 = (a – b) (a2 + ab + b2),
where,
- a = 25
- b = 12
= 253 – 123
= (25 – 12) (252 + (25)(12) + 122)
= 13 (625 + 300 + 144)
= 13897
Example 3: Factorize 8p3 + 27
Solution:
8p3 + 27 = (2p)3 + (3)3
Since, a3 + b3 = (a + b)(a2 – ab + b2)
= (2p)3 + (3)3
= (2p + 3)[(2p)2 – (2p)(3) + (3)2]
= (2p + 3)[4p2 – 6p + 9]
Example 4: Factorize 512 + 729v3
Solution:
512 + 729v3 = (8)3 + (9v)3
Since, a3 + b3 = (a + b)(a2 – ab + b2)
= (8)3 + (9v)3
= (8 + 9v)[(8)2 – (8)(9v) + (9v)2]
= (8 + 9v)[64 – 72v + 729v2]
Example 5: Solve: 253 + 123
Solution:
Since, a3 + b3 = (a + b) (a2 – ab + b2)
where,
- a = 25
- b = 12
= 253 + 123
= (25 + 12) (252 - (25)(12) + 122)
= 37 (625 - 300 + 144)
= 17353
Practice Problems on Sum and Differences of Cubes
Q1. Factorize 64 + 343v3
Q2. Factorize 64 - 343v3
Q3. Evaluate 153 – 93
Q4. Evaluate 233 – 73