(a + b)³ and (a - b)³ Identities
The cube identities (a + b)³ and (a − b)³ are algebraic identities used to expand or factorise expressions involving cubes. They are extensions of the square identities studied in Class 8.
In NCERT Class 9 Mathematics (Chapter 2 — Polynomials), these identities are introduced alongside other polynomial identities. They are essential for factorisation, simplification, and solving algebraic problems.
The cube identities appear frequently in competitive examinations, higher algebra, and applied mathematics. Mastering them enables quick computation of expressions like 21³, 99³, and similar values without lengthy multiplication.
These identities are also connected to the sum and difference of cubes (a³ + b³ and a³ − b³), which are studied as separate identities.
What is (a + b)³ and (a − b)³ Identities?
The two cube identities:
Identity 1: (a + b)³
(a + b)³ = a³ + 3a²b + 3ab² + b³
This can also be written as:
(a + b)³ = a³ + b³ + 3ab(a + b)
Identity 2: (a − b)³
(a − b)³ = a³ − 3a²b + 3ab² − b³
This can also be written as:
(a − b)³ = a³ − b³ − 3ab(a − b)
Key observations:
- Both identities have four terms when fully expanded.
- The signs alternate in (a − b)³: +, −, +, −.
- All signs are positive in (a + b)³.
- The middle two terms can be combined using 3ab(a ± b).
(a + b)³ and (a - b)³ Identities Formula
Key Formulas:
1. (a + b)³ expanded:
a³ + 3a²b + 3ab² + b³
2. (a − b)³ expanded:
a³ − 3a²b + 3ab² − b³
3. Compact forms:
- (a + b)³ = a³ + b³ + 3ab(a + b)
- (a − b)³ = a³ − b³ − 3ab(a − b)
4. Derived results:
- a³ + b³ = (a + b)³ − 3ab(a + b)
- a³ − b³ = (a − b)³ + 3ab(a − b)
5. Useful for computation:
- If a + b = s and ab = p, then (a + b)³ = s³ and a³ + b³ = s³ − 3ps.
6. Connection to Pascal’s triangle:
- Coefficients of (a + b)³: 1, 3, 3, 1 (row 3 of Pascal’s triangle).
- Powers of a decrease: a³, a², a, 1.
- Powers of b increase: 1, b, b², b³.
Derivation and Proof
Derivation of (a + b)³:
Method: Multiplying (a + b) by itself three times.
Step 1: Start with (a + b)² = a² + 2ab + b² (known identity).
Step 2: Multiply (a + b)² by (a + b):
- (a + b)³ = (a + b)² × (a + b)
- = (a² + 2ab + b²)(a + b)
- = a²(a + b) + 2ab(a + b) + b²(a + b)
- = a³ + a²b + 2a²b + 2ab² + ab² + b³
- = a³ + 3a²b + 3ab² + b³
Therefore: (a + b)³ = a³ + 3a²b + 3ab² + b³
Rearranging the middle terms:
- 3a²b + 3ab² = 3ab(a + b)
- So: (a + b)³ = a³ + b³ + 3ab(a + b)
Derivation of (a − b)³:
Replace b with (−b) in the (a + b)³ identity:
- (a + (−b))³ = a³ + 3a²(−b) + 3a(−b)² + (−b)³
- = a³ − 3a²b + 3ab² − b³
Therefore: (a − b)³ = a³ − 3a²b + 3ab² − b³
Types and Properties
Common applications and problem types:
1. Direct expansion
- Expand (2x + 3y)³ or (p − 2q)³ using the identities.
- Substitute a and b with the given terms and apply the formula.
2. Quick computation of cubes
- Find 21³ by writing it as (20 + 1)³.
- Find 99³ by writing it as (100 − 1)³.
3. Factorisation
- Recognise expressions in the form a³ + 3a²b + 3ab² + b³ and write them as (a + b)³.
4. Finding a³ + b³ or a³ − b³
- Given a + b and ab, find a³ + b³ using: a³ + b³ = (a + b)³ − 3ab(a + b).
5. Simplification
- Simplify expressions like (x + 2)³ − x³ by expanding and cancelling.
6. Verification
- Verify the identity by substituting specific values of a and b.
Solved Examples
Example 1: Example 1: Expanding (a + b)³
Problem: Expand (2x + 3)³.
Solution:
Using (a + b)³ = a³ + 3a²b + 3ab² + b³:
- a = 2x, b = 3
- = (2x)³ + 3(2x)²(3) + 3(2x)(3)² + (3)³
- = 8x³ + 3(4x²)(3) + 3(2x)(9) + 27
- = 8x³ + 36x² + 54x + 27
Answer: (2x + 3)³ = 8x³ + 36x² + 54x + 27.
Example 2: Example 2: Expanding (a − b)³
Problem: Expand (3p − 2q)³.
Solution:
Using (a − b)³ = a³ − 3a²b + 3ab² − b³:
- a = 3p, b = 2q
- = (3p)³ − 3(3p)²(2q) + 3(3p)(2q)² − (2q)³
- = 27p³ − 3(9p²)(2q) + 3(3p)(4q²) − 8q³
- = 27p³ − 54p²q + 36pq² − 8q³
Answer: (3p − 2q)³ = 27p³ − 54p²q + 36pq² − 8q³.
Example 3: Example 3: Quick computation of 21³
Problem: Find 21³ using the identity.
Solution:
Write 21 = 20 + 1:
- (20 + 1)³ = 20³ + 3(20)²(1) + 3(20)(1)² + 1³
- = 8000 + 3(400) + 3(20) + 1
- = 8000 + 1200 + 60 + 1
- = 9261
Answer: 21³ = 9261.
Example 4: Example 4: Quick computation of 99³
Problem: Find 99³ using the identity.
Solution:
Write 99 = 100 − 1:
- (100 − 1)³ = 100³ − 3(100)²(1) + 3(100)(1)² − 1³
- = 1000000 − 30000 + 300 − 1
- = 970299
Answer: 99³ = 970299.
Example 5: Example 5: Finding a³ + b³ given a + b and ab
Problem: If a + b = 5 and ab = 6, find a³ + b³.
Solution:
Using a³ + b³ = (a + b)³ − 3ab(a + b):
- = 5³ − 3(6)(5)
- = 125 − 90
- = 35
Answer: a³ + b³ = 35.
Example 6: Example 6: Finding a³ − b³ given a − b and ab
Problem: If a − b = 3 and ab = 10, find a³ − b³.
Solution:
Using a³ − b³ = (a − b)³ + 3ab(a − b):
- = 3³ + 3(10)(3)
- = 27 + 90
- = 117
Answer: a³ − b³ = 117.
Example 7: Example 7: Factorising an expression
Problem: Factorise: 8x³ + 12x² + 6x + 1.
Solution:
Recognise the pattern:
- 8x³ = (2x)³
- 1 = 1³
- 12x² = 3(2x)²(1)
- 6x = 3(2x)(1)²
This matches (a + b)³ = a³ + 3a²b + 3ab² + b³ with a = 2x, b = 1.
Answer: 8x³ + 12x² + 6x + 1 = (2x + 1)³.
Example 8: Example 8: Factorising a difference cube pattern
Problem: Factorise: 27a³ − 54a²b + 36ab² − 8b³.
Solution:
Recognise the pattern:
- 27a³ = (3a)³
- 8b³ = (2b)³
- 54a²b = 3(3a)²(2b)
- 36ab² = 3(3a)(2b)²
This matches (a − b)³ with a = 3a, b = 2b.
Answer: = (3a − 2b)³.
Example 9: Example 9: Verifying the identity
Problem: Verify (a + b)³ = a³ + 3a²b + 3ab² + b³ for a = 2, b = 3.
Solution:
LHS:
- (2 + 3)³ = 5³ = 125
RHS:
- 2³ + 3(2)²(3) + 3(2)(3)² + 3³
- = 8 + 36 + 54 + 27 = 125
LHS = RHS = 125 ✓
Answer: Identity verified.
Example 10: Example 10: Simplification problem
Problem: Simplify: (x + 2)³ − (x − 2)³.
Solution:
Expand both cubes:
- (x + 2)³ = x³ + 3x²(2) + 3x(4) + 8 = x³ + 6x² + 12x + 8
- (x − 2)³ = x³ − 3x²(2) + 3x(4) − 8 = x³ − 6x² + 12x − 8
Subtract:
- = (x³ + 6x² + 12x + 8) − (x³ − 6x² + 12x − 8)
- = x³ + 6x² + 12x + 8 − x³ + 6x² − 12x + 8
- = 12x² + 16
Answer: (x + 2)³ − (x − 2)³ = 12x² + 16.
Real-World Applications
Applications of Cube Identities:
- Quick computation: Calculate cubes of numbers close to round numbers (21³, 99³, 52³) without lengthy multiplication.
- Algebraic factorisation: Recognise and factorise expressions that match the cube identity pattern.
- Simplification: Simplify complex algebraic expressions involving cubes.
- Finding a³ + b³ or a³ − b³: Use the rearranged forms to find cube sums/differences given a + b and ab.
- Physics and engineering: Cube relationships appear in volume calculations, scaling laws, and power formulas.
- Higher mathematics: The binomial theorem (generalisation of these identities to any power) is built on the patterns seen here.
Key Points to Remember
- (a + b)³ = a³ + 3a²b + 3ab² + b³
- (a − b)³ = a³ − 3a²b + 3ab² − b³
- Compact form: (a + b)³ = a³ + b³ + 3ab(a + b).
- Compact form: (a − b)³ = a³ − b³ − 3ab(a − b).
- The coefficients follow Pascal’s triangle row 3: 1, 3, 3, 1.
- Signs in (a − b)³ alternate: +, −, +, −.
- a³ + b³ = (a + b)³ − 3ab(a + b) — useful for finding cube sums.
- a³ − b³ = (a − b)³ + 3ab(a − b) — useful for finding cube differences.
- These identities are used for factorisation, simplification, and quick computation.
- Derived by multiplying (a ± b)² by (a ± b) using the distributive law.
Practice Problems
- Expand (x + 4)³.
- Expand (2a − 5b)³.
- Find 31³ using the identity (30 + 1)³.
- Find 48³ using the identity (50 − 2)³.
- If x + y = 7 and xy = 12, find x³ + y³.
- If p − q = 4 and pq = 5, find p³ − q³.
- Factorise: 125x³ + 75x² + 15x + 1.
- Simplify: (a + 1)³ − (a − 1)³.
Frequently Asked Questions
Q1. What is the formula for (a + b)³?
(a + b)³ = a³ + 3a²b + 3ab² + b³. It can also be written as a³ + b³ + 3ab(a + b).
Q2. What is the formula for (a − b)³?
(a − b)³ = a³ − 3a²b + 3ab² − b³. It can also be written as a³ − b³ − 3ab(a − b).
Q3. How do you remember the signs in (a − b)³?
The signs alternate: +, −, +, −. The first and third terms are positive (a³ and 3ab²), and the second and fourth are negative (−3a²b and −b³).
Q4. How are these identities derived?
By multiplying (a + b)² by (a + b). First expand (a + b)² = a² + 2ab + b², then multiply by (a + b) using the distributive law.
Q5. How do you find a³ + b³ if a + b and ab are given?
Use: a³ + b³ = (a + b)³ − 3ab(a + b). Substitute the known values of (a + b) and ab.
Q6. What are the coefficients in (a + b)³?
The coefficients are 1, 3, 3, 1 — the fourth row of Pascal's triangle. They correspond to the terms a³, a²b, ab², b³.
Q7. How do you use these identities to compute 101³?
Write 101 = 100 + 1. Then (100 + 1)³ = 100³ + 3(100)²(1) + 3(100)(1) + 1 = 1000000 + 30000 + 300 + 1 = 1030301.
Q8. Are (a + b)³ and (a − b)³ in the CBSE Class 9 syllabus?
Yes. These identities are part of Chapter 2 (Polynomials) in the CBSE Class 9 Mathematics syllabus, used for expansion and factorisation.
Q9. What is the difference between (a + b)³ and a³ + b³?
(a + b)³ includes the cross terms 3a²b and 3ab², while a³ + b³ is just the sum of cubes. The relationship is: a³ + b³ = (a + b)³ − 3ab(a + b).
Q10. Can you factorise 8x³ + 12x² + 6x + 1?
Yes. Recognise it as (2x)³ + 3(2x)²(1) + 3(2x)(1)² + 1³ = (2x + 1)³.
Related Topics
- Algebraic Identities (Extended)
- a³ + b³ and a³ - b³ Identities
- (a + b)² Identity
- Factorisation of Polynomials
- Polynomials in One Variable
- Degree of a Polynomial
- Types of Polynomials
- Value of a Polynomial
- Zeroes of a Polynomial
- Remainder Theorem
- Factor Theorem
- Zeroes of Quadratic Polynomial
- Relationship Between Zeroes and Coefficients
- Sum and Product of Zeroes
