Algebraic Identities Chapter Notes | Mathematics Class 8 ICSE PDF Download

Introduction

In mathematics, identities are powerful tools that help us simplify and solve algebraic expressions efficiently. This chapter introduces the concept of special products and expansions, which are shortcuts to multiply and expand expressions. By understanding these identities, you can quickly perform calculations involving binomials, trinomials, and higher powers. The chapter covers special products, the product of sum and difference, expansions of squares and cubes, and their applications in solving problems, making algebraic manipulations easier and more efficient.

Special Products

• Special products are quick methods to multiply specific types of expressions, especially binomials.
• These methods help simplify multiplication by recognizing patterns in the results.
• Steps for multiplying two binomials (x + a)(x + b):
  • Multiply the first terms: x × x = x².
  • Add the products of outer and inner terms: x × b + a × x = bx + ax.
  • Multiply the second terms: a × b = ab.
  • Combine: x² + (a + b)x + ab.
• Key identities:
  • (x + a)(x + b) = x² + (a + b)x + ab
  • (x + a)(x - b) = x² + (a - b)x - ab
  • (x - a)(x + b) = x² + (b - a)x - ab
  • (x - a)(x - b) = x² - (a + b)x + ab

Example: Multiply (x + 4)(x + 2).
Using the identity (x + a)(x + b) = x² + (a + b)x + ab, where a = 4, b = 2:

• First term: x²
• Middle term: (4 + 2)x = 6x
• Last term: 4 × 2 = 8
• Result: x² + 6x + 8

Important

• When multiplying two binomials using the direct method, the result has three terms.
• Steps to find the product:
  • First term: Multiply the first terms of both binomials.
  • Middle term: Add the product of outer terms and inner terms.
  • Third term: Multiply the second terms of both binomials.

Example: Multiply (3x + 2y)(4x + 5y).
First term: 3x × 4x = 12x²
Middle term: (3x × 5y) + (2y × 4x) = 15xy + 8xy = 23xy
Third term: 2y × 5y = 10y²
Result: 12x² + 23xy + 10y²

Product of Sum and Difference of Two Terms

• The product of the sum and difference of two terms follows the identity: (x + y)(x - y) = x² - y².
• This is called the difference of squares.
• Steps:
  • Identify the two terms (e.g., x and y).
  • Compute the square of the first term (x²).
  • Subtract the square of the second term (y²).
• This identity can also be used to simplify numerical multiplications.

Example: Evaluate 105 × 95.
Rewrite as (100 + 5)(100 - 5).
Using (x + y)(x - y) = x² - y², where x = 100, y = 5:
- x² = 100² = 10000
- y² = 5² = 25
Result: 10000 - 25 = 9975

Expansions

• Expansion involves multiplying an expression by itself to find its square or higher powers.
• Key identities for squaring:
  • (a + b)² = a² + 2ab + b²
  • (a - b)² = a² - 2ab + b²
• Steps for (a + b)²:
  • Square the first term: a².
  • Add twice the product of the terms: 2ab.
  • Add the square of the second term: b².

Example: Expand (2x + 3y)².
Using (a + b)² = a² + 2ab + b², where a = 2x, b = 3y:
- a² = (2x)² = 4x²
- 2ab = 2 × 2x × 3y = 12xy
- b² = (3y)² = 9y²
Result: 4x² + 12xy + 9y²

Important Formulae to be Memorised

Memorize these key identities for quick calculations:

• (a + b)² = a² + b² + 2ab
• (a - b)² = a² + b² - 2ab
• (a + 1/a)² = a² + 1/a² + 2
• (a - 1/a)² = a² + 1/a² - 2
• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
• (a + b - c)² = a² + b² + c² + 2ab - 2bc - 2ca

Example: Expand (a + 2b - 3c)².
Using (a + b - c)² = a² + b² + c² + 2ab - 2bc - 2ca, where a = a, b = 2b, c = 3c:
- a² = a²
- b² = (2b)² = 4b²
- c² = (3c)² = 9c²
- 2ab = 2 × a × 2b = 4ab
- 2bc = 2 × 2b × 3c = 12bc
- 2ca = 2 × 3c × a = 6ca
Result: a² + 4b² + 9c² + 4ab - 12bc - 6ca

Cubes of Binomials

• Cubing a binomial means multiplying it by itself three times.
• Key identities:
  • (a + b)³ = a³ + 3a²b + 3ab² + b³
  • (a - b)³ = a³ - 3a²b + 3ab² - b³
• Alternative forms:
  • (a + b)³ = a³ + b³ + 3ab(a + b)
  • (a - b)³ = a³ - b³ - 3ab(a - b)

Example: Expand (2x + y)³.
Using (a + b)³ = a³ + 3a²b + 3ab² + b³, where a = 2x, b = y:
- a³ = (2x)³ = 8x³
- 3a²b = 3 × (2x)² × y = 3 × 4x² × y = 12x²y
- 3ab² = 3 × 2x × y² = 6xy²
- b³ = y³
Result: 8x³ + 12x²y + 6xy² + y³

Application of Formulae

• Identities can be used to find sums, differences, or products of terms when given specific values.
• Steps for using identities like (a + b)² or (a + b)³:
  • Substitute the given values into the identity.
  • Perform the calculations step-by-step.
  • Simplify to find the required expression.
• These identities help solve problems involving squares, cubes, or sums of terms efficiently.

Example: If a + b = 7 and ab = 12, find a² + b².
Using (a + b)² = a² + b² + 2ab:
- (a + b)² = 7² = 49
- 2ab = 2 × 12 = 24
- a² + b² = 49 - 24 = 25
Result: a² + b² = 25

Example : If a + b = 5 and ab = 6, find a³ + b³.

Solution :
(a + b)³ = a³ + b³ + 3ab(a + b)
=> (5)³ = a³ + b³ + 3 × 6 × 5
=> 125 = a³ + b³ + 90
=> a³ + b³ = 35

Example: If a - 1/a = 3; find a³ - 1/a³

Solution :
Since, (a - 1/a)³ = a³ - 1/a³ - 3a(a - 1/a)

(a - 1/a)³ = a³ - 1/a³ - 3a(a - 1/a)
(3)³ = a³ - 1/a³ - 3 × 3 × 3
=> 27 = a³ - 1/a³ - 27
=> a³ - 1/a³ = 54

Example :
The sum of two numbers is 4 and their product is 3. Find :
(i) the sum of their squares.
(ii) the sum of their cubes.

Solution :
Let the numbers be x and y.
Therefore x + y = 4 and xy = 3

To find (i) x² + y² (ii) x³ + y³
(i) x² + y² = x² + y² + 2xy
(x + y)² = x² + y² + 2xy
4² = x² + y² + 2 × 3
16 = x² + y² + 6
x² + y² = 10

(ii) (x + y)³ = x³ + y³ + 3xy(x + y)
4³ = x³ + y³ + 3 × 3 × 4
64 = x³ + y³ + 36
x³ + y³ = 28