Exploring Algebraic Identities

Algebraic Identities: Complete CBSE Class 9 Notes

Definition and Introduction to Algebraic Identities

An **algebraic identity** is an equation that is true for **all values** of the variables occurring in it. This is different from an algebraic equation, which is true only for specific values of variables.

**Example of equation:** x² – 1 = 24 is true only when x = 5 or x = –5

**Example of identity:** (x + y)² = x² + 2xy + y² is true for all values of x and y

**Key distinction:** Identities are universal truths in algebra, while equations have limited solutions.

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Identity 1: (a + b)² = a² + 2ab + b²

Geometric Visualization

A square of side (a + b) units can be divided into four parts:

One square of side a with area a²

One square of side b with area b²

Two rectangles of dimensions a × b with total area 2ab

Therefore: **(a + b)² = a² + 2ab + b²**

Algebraic Proof Using Distributive Property

(a + b)² = (a + b)(a + b)

= a(a + b) + b(a + b)

= a² + ab + ba + b²

= **a² + 2ab + b²**

Validity for All Number Types

This identity holds for:

Positive numbers: If a = 10, b = 2, then (12)² = 100 + 40 + 4 = 144 ✓

Negative numbers: If a = –2, b = –3, then (–5)² = 4 + 12 + 9 = 25 ✓

Rational numbers: Works for all fractions

Applications

**Expansion of binomials:**

(5x + 2y)² = (5x)² + 2(5x)(2y) + (2y)² = 25x² + 20xy + 4y²

**Numerical calculations:**

43² = (40 + 3)² = 1600 + 240 + 9 = 1849

**Important pattern observation:**

(a + b)² ≠ a² + b²

(a + b)² = a² + b² + 2ab

(a + b)² is always greater than a² + b² (when ab > 0)

The difference is determined by the term 2ab

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Identity 2: (a – b)² = a² – 2ab + b²

Geometric Visualization

In a square of side a units, if we remove a smaller section of side b, the remaining area is (a – b)². This can be decomposed as:

Area of large square: a²

Minus two rectangles of dimensions a × b and b × (a – b)

Plus overlap correction: gives us (a – b)² = a² – 2ab + b²

Algebraic Proof

(a – b)² = (a – b)(a – b)

= a(a – b) – b(a – b)

= a² – ab – ba + b²

= **a² – 2ab + b²**

Comparison with (a + b)²

The only difference is the **sign of the middle term**:

(a + b)²: coefficient is **+2ab**

(a – b)²: coefficient is **–2ab**

Applications

**Expansion:**

(7x – 3y)² = 49x² – 42xy + 9y²

**Numerical calculations:**

29² = (30 – 1)² = 900 – 60 + 1 = 841

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Identity 3: (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Derivation

Starting with (a + b + c)², replace (b + c) with d:

(a + d)² = a² + 2ad + d²

Substitute back d = (b + c):

= a² + 2a(b + c) + (b + c)²

= a² + 2ab + 2ac + b² + 2bc + c²

**Rearranged form:** (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Memory Aid

Terms with squares: a², b², c²

Terms with products: 2ab + 2bc + 2ca (each pair of variables multiplied by 2)

Geometric Interpretation

A square of side (a + b + c) contains:

Three small squares of areas a², b², c²

Six rectangles of areas ab, ab, bc, bc, ca, ca

Total: a² + b² + c² + 2(ab + bc + ca)

Application Example

119² = (100 + 10 + 9)²

= 10000 + 100 + 81 + 2(1000) + 2(900) + 2(90)

= 10000 + 100 + 81 + 2000 + 1800 + 180

= **14161**

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Identity 4: a² – b² = (a + b)(a – b)

Proof

(a + b)(a – b) = a(a – b) + b(a – b)

= a² – ab + ba – b²

= **a² – b²**

Alternative Form (Śhrīdharāchārya's Method, 750 CE)

a² = (a + b)(a – b) + b²

This can be rearranged to: **(a + b)(a – b) = a² – b²**

Application for Quick Calculations

**Example:** 55² = (55 + 5)(55 – 5) + 5²

= 60 × 50 + 25

= 3000 + 25

= **3025**

**General pattern for numbers ending in 5:**

35² = 40 × 30 + 25 = 1200 + 25 = 1225

65² = 70 × 60 + 25 = 4200 + 25 = 4225

85² = 90 × 80 + 25 = 7200 + 25 = 7225

**Interesting observation:** Numbers ending in 5 always have squares ending in 25

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Factorization Using (a + b)² = a² + 2ab + b²

Procedure

To factor an expression of form a² + 2ab + b²:

1. Identify if the first and last terms are perfect squares

2. Check if the middle term equals 2 × (square root of first term) × (square root of last term)

3. If yes, factor as (a + b)²

Example 1: x² + 4x + 4

First term: x² = (x)²

Last term: 4 = (2)²

Middle term: 4x = 2(x)(2) ✓

**Factor:** (x + 2)²

Example 2: 36x² + 12x + 1

First term: 36x² = (6x)²

Last term: 1 = (1)²

Middle term: 12x = 2(6x)(1) ✓

**Factor:** (6x + 1)²

Example 3: 50p² + 60pq + 18q²

Step 1: Extract common factor: 2(25p² + 30pq + 9q²)

Step 2: Factor the trinomial:

(25p²) = (5p)²

(9q²) = (3q)²

30pq = 2(5p)(3q) ✓

**Factor:** 2(5p + 3q)²

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Factorization Using (a – b)² = a² – 2ab + b²

Procedure

To factor an expression of form a² – 2ab + b²:

1. Identify if first and last terms are perfect squares

2. Check if middle term = –2 × (√first term) × (√last term)

3. If yes, factor as (a – b)²

Example: 16y² – 24y + 9

First term: 16y² = (4y)²

Last term: 9 = (3)²

Middle term: –24y = –2(4y)(3) ✓

**Factor:** (4y – 3)²

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Proof of Pattern from Example 1

**Pattern:** For any three consecutive square numbers, if you add the smallest and largest, then subtract twice the middle square, the result is always 2.

**Proof using identities:**

Any three consecutive integers: (n – 1), n, (n + 1)

Their squares: (n – 1)², n², (n + 1)²

Sum of smallest and largest:

(n – 1)² + (n + 1)² = (n² – 2n + 1) + (n² + 2n + 1) = 2n² + 2

Subtract twice the middle square:

2n² + 2 – 2n² = **2** ✓

This proves the pattern algebraically using the (a ± b)² identities.

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Factorization of Quadratic Expressions Using Algebra Tiles

Concept

For expression x² + (a + b)x + ab, the factorization is (x + a)(x + b)

**Condition:** We must find two numbers whose:

Sum = coefficient of x

Product = constant term

Example: x² + 7x + 12

**Find a and b such that:**

a + b = 7

ab = 12

**Solution:** a = 3, b = 4

**Factorization:** (x + 3)(x + 4)

**Verification:** (x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12 ✓

Example: x² + 11x + 30

**Conditions:**

a + b = 11

ab = 30

**Factors of 30:** 1×30, 2×15, 3×10, 5×6

**Check:** 5 + 6 = 11 and 5 × 6 = 30 ✓

**Factorization:** (x + 5)(x + 6)

Example: x² – 5x + 6

**Conditions:**

a + b = –5

ab = 6

**Solution:** a = –2, b = –3 (both negative)

**Factorization:** (x – 2)(x – 3)

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Factorization of Trinomials ax² + bx + c (where a ≠ 1)

General Form

(px + a)(qx + b) = pqx² + (pb + qa)x + ab

Example: 6x² + 7x + 2

**Method:**

1. Find two numbers whose product = 6 × 2 = 12 and sum = 7

2. These are 3 and 4

3. Split: 6x² + 3x + 4x + 2

4. Group: 3x(2x + 1) + 2(2x + 1)

5. **Factor:** (3x + 2)(2x + 1)

**Verification:** (3x + 2)(2x + 1) = 6x² + 3x + 4x + 2 = 6x² + 7x + 2 ✓

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Identity 5: (a + b)³ = a³ + 3a²b + 3ab² + b³

Derivation

(a + b)³ = (a + b)(a + b)²

= (a + b)(a² + 2ab + b²)

Expand using distributive property:

= a(a² + 2ab + b²) + b(a² + 2ab + b²)

= a³ + 2a²b + ab² + a²b + 2ab² + b³

= **a³ + 3a²b + 3ab² + b³**

Geometric Visualization

A cube of edge (a + b) can be divided into:

One cube of edge a: volume = a³

One cube of edge b: volume = b³

Three cuboids of dimensions a × a × b: volume = 3a²b

Three cuboids of dimensions a × b × b: volume = 3ab²

**Total:** (a + b)³ = a³ + 3a²b + 3ab² + b³

Pattern in Coefficients

Coefficients: 1, 3, 3, 1 (binomial coefficients from Pascal's triangle)

Alternate powers of a and b decrease and increase respectively

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Identity 6: (a – b)³ = a³ – 3a²b + 3ab² – b³

Derivation

Replace b with (–b) in (a + b)³:

(a – b)³ = a³ + 3a²(–b) + 3a(–b)² + (–b)³

= **a³ – 3a²b + 3ab² – b³**

Sign Pattern

1st term: positive (a³)

2nd term: negative (–3a²b)

3rd term: positive (3ab²)

4th term: negative (–b³)

Signs **alternate** starting with positive

Example

(2x – 1)³ = (2x)³ – 3(2x)²(1) + 3(2x)(1)² – (1)³

= 8x³ – 12x² + 6x – 1

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Application: Finding Cube Roots and Sides of Cubes

Example 1: p³ + 6p²q + 12pq² + 8q³

**Match with (a + b)³ = a³ + 3a²b + 3ab² + b³:**

a³ = p³ → a = p

b³ = 8q³ → b = 2q

Check: 3a²b = 3p²(2q) = 6p²q ✓

Check: 3ab² = 3p(2q)² = 12pq² ✓

**Side of cube:** (p + 2q) units

Example 2: 8n³ – 60n²m + 150nm² – 125m³

**Match with (a – b)³ = a³ – 3a²b + 3ab² – b³:**

a³ = 8n³ → a = 2n

b³ = 125m³ → b = 5m

Check: –3a²b = –3(2n)²(5m) = –60n²m ✓

Check: 3ab² = 3(2n)(5m)² = 150nm² ✓

**Expression:** (2n – 5m)³

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Key Exam Tips and Common Mistakes

**1. Don't confuse (a + b)² with a² + b²**

(a + b)² = a² + 2ab + b² (middle term is crucial)

Always expand fully before substituting values

**2. Sign care in (a – b)²**

Middle term is **negative**: –2ab

When factoring, both factors must have same sign pattern

**3. Factorization procedure**

Always check for common factors first (GCF)

Then apply appropriate identity

Verify by expanding the factors

**4. For ax² + bx + c (a ≠ 1)**

Find numbers whose product = a × c and sum = b

This method works for all quadratic trinomials

**5. Cube identities**

Memorize the alternating sign pattern for (a – b)³

Coefficients are always 1, 3, 3, 1

**6. Quick calculations**

Use a² – b² for products like 41 × 39 = (40 + 1)(40 – 1) = 1600 – 1 = 1599

Use (a ± b)² for squaring numbers close to multiples of 10

Use (a + b + c)² when number splits into three parts conveniently

**7. Verification**

Always expand factored form to verify

For identities, test with specific values as check

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Summary Table of All Identities

| Identity | Form | Use |

|----------|------|-----|

| (a+b)² | a² + 2ab + b² | Expansion, squaring |

| (a–b)² | a² – 2ab + b² | Factorization |

| (a+b+c)² | a² + b² + c² + 2ab + 2bc + 2ca | Three-term squaring |

| a²–b² | (a+b)(a–b) | Difference of squares |

| (a+b)³ | a³ + 3a²b + 3ab² + b³ | Cubic expansion |

| (a–b)³ | a³ – 3a²b + 3ab² – b³ | Cubic factorization |

These identities form the foundation for advanced algebraic manipulation and are essential for CBSE board examinations.