Simplified Quantitative Formulas: Algebraic Identities
- Identity: An equation true for all values of variables. E.g., (a+b)² = a²+2ab+b².
- Basic: a+b=b+a, a×b=b×a, a(b+c)=ab+ac.
- Squares: (a+b)², (a-b)², a²-b², (a+b+c)².
- Cubes: (a+b)³, (a-b)³, a³+b³, a³-b³.
- Special: (x+a)(x+b)=x²+(a+b)x+ab, etc.
What do these mean? (Super Simple Explanations & Examples)
- Identity: Always true, no matter what values you use.
- Example: (a+b)² = a²+2ab+b². Try a=2, b=3: (2+3)²=25, 2²+2×2×3+3²=25. - Square: (a-b)² = a²-2ab+b². Example: a=4, b=1: (4-1)²=9, 4²-2×4×1+1²=9.
- Cube: (a+b)³ = a³+3a²b+3ab²+b³. Example: a=1, b=2: (1+2)³=27, 1³+3×1²×2+3×1×2²+2³=27.
- Special: (x+a)(x+b)=x²+(a+b)x+ab. Example: x=2, a=1, b=3: (2+1)(2+3)=3×5=15, 2²+3×2+3=15.
Introduction to Algebraic Identities
Algebraic identities are equations that are true for all values of the variables. They are fundamental tools in algebra and are used extensively in solving equations and simplifying expressions.
What are Algebraic Identities?
An algebraic identity is an equation that holds true for all values of the variables involved. Unlike equations, identities are always true regardless of the values substituted for the variables.
Example 1
Verify that (a + b)² = a² + 2ab + b² is an identity by substituting different values for a and b.
Solution:
Let's try a = 2, b = 3:
LHS: (2 + 3)² = 5² = 25
RHS: 2² + 2(2)(3) + 3² = 4 + 12 + 9 = 25
Let's try a = -1, b = 4:
LHS: (-1 + 4)² = 3² = 9
RHS: (-1)² + 2(-1)(4) + 4² = 1 - 8 + 16 = 9
The identity holds true for all values of a and b.
Basic Algebraic Identities
These are the fundamental identities that form the basis for more complex ones.
Basic Identities
- a + b = b + a (Commutative Law of Addition)
- a × b = b × a (Commutative Law of Multiplication)
- (a + b) + c = a + (b + c) (Associative Law of Addition)
- (a × b) × c = a × (b × c) (Associative Law of Multiplication)
- a(b + c) = ab + ac (Distributive Law)
Example 2
Simplify: 3(x + 2) + 4(x - 1)
Solution:
Using distributive law:
3(x + 2) + 4(x - 1)
= 3x + 6 + 4x - 4
= (3x + 4x) + (6 - 4)
= 7x + 2
Square Identities
These identities involve squares of expressions.
Square Identities with Proofs
1. (a + b)² = a² + 2ab + b²
Proof:
(a + b)² = (a + b)(a + b)
= a(a + b) + b(a + b)
= a² + ab + ba + b²
= a² + 2ab + b²
This identity shows that the square of a sum is equal to the sum of squares plus twice the product.
2. (a - b)² = a² - 2ab + b²
Proof:
(a - b)² = (a - b)(a - b)
= a(a - b) - b(a - b)
= a² - ab - ba + b²
= a² - 2ab + b²
This identity shows that the square of a difference is equal to the sum of squares minus twice the product.
3. a² - b² = (a + b)(a - b)
Proof:
(a + b)(a - b) = a(a - b) + b(a - b)
= a² - ab + ba - b²
= a² - b²
This identity shows that the difference of squares can be factored into the product of sum and difference.
4. (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Proof:
(a + b + c)² = [(a + b) + c]²
= (a + b)² + 2(a + b)c + c²
= a² + 2ab + b² + 2ac + 2bc + c²
= a² + b² + c² + 2ab + 2bc + 2ca
This identity shows that the square of a trinomial is equal to the sum of squares of all terms plus twice the product of each pair of terms.
Example 3
Expand: (2x + 3y)²
Solution:
Using (a + b)² = a² + 2ab + b²:
(2x + 3y)² = (2x)² + 2(2x)(3y) + (3y)²
= 4x² + 12xy + 9y²
Example 4
Factorize: 9x² - 16y²
Solution:
Using a² - b² = (a + b)(a - b):
9x² - 16y² = (3x)² - (4y)²
= (3x + 4y)(3x - 4y)
Cube Identities
These identities involve cubes of expressions.
Cube Identities with Proofs
1. (a + b)³ = a³ + 3a²b + 3ab² + b³
Proof:
(a + b)³ = (a + b)(a + b)²
= (a + b)(a² + 2ab + b²)
= a(a² + 2ab + b²) + b(a² + 2ab + b²)
= a³ + 2a²b + ab² + a²b + 2ab² + b³
= a³ + 3a²b + 3ab² + b³
This identity shows that the cube of a sum is equal to the sum of cubes plus three times the product of square and linear terms.
2. (a - b)³ = a³ - 3a²b + 3ab² - b³
Proof:
(a - b)³ = (a - b)(a - b)²
= (a - b)(a² - 2ab + b²)
= a(a² - 2ab + b²) - b(a² - 2ab + b²)
= a³ - 2a²b + ab² - a²b + 2ab² - b³
= a³ - 3a²b + 3ab² - b³
This identity shows that the cube of a difference is equal to the difference of cubes minus three times the product of square and linear terms.
3. a³ + b³ = (a + b)(a² - ab + b²)
Proof:
(a + b)(a² - ab + b²) = a(a² - ab + b²) + b(a² - ab + b²)
= a³ - a²b + ab² + a²b - ab² + b³
= a³ + b³
This identity shows that the sum of cubes can be factored into the product of sum and a specific quadratic expression.
4. a³ - b³ = (a - b)(a² + ab + b²)
Proof:
(a - b)(a² + ab + b²) = a(a² + ab + b²) - b(a² + ab + b²)
= a³ + a²b + ab² - a²b - ab² - b³
= a³ - b³
This identity shows that the difference of cubes can be factored into the product of difference and a specific quadratic expression.
Example 5
Expand: (x + 2)³
Solution:
Using (a + b)³ = a³ + 3a²b + 3ab² + b³:
(x + 2)³ = x³ + 3x²(2) + 3x(2)² + 2³
= x³ + 6x² + 12x + 8
Example 6
Factorize: x³ + 8
Solution:
Using a³ + b³ = (a + b)(a² - ab + b²):
x³ + 8 = x³ + 2³
= (x + 2)(x² - 2x + 4)
Special Identities
These are additional useful identities for specific cases.
Special Identities with Proofs
1. (a + b + c)³ = a³ + b³ + c³ + 3(a + b)(b + c)(c + a)
Proof:
(a + b + c)³ = [(a + b) + c]³
= (a + b)³ + 3(a + b)²c + 3(a + b)c² + c³
= a³ + 3a²b + 3ab² + b³ + 3(a² + 2ab + b²)c + 3(a + b)c² + c³
= a³ + b³ + c³ + 3(a + b)(b + c)(c + a)
This identity shows the expansion of a trinomial cube in terms of individual cubes and products of pairs.
2. a⁴ - b⁴ = (a² + b²)(a + b)(a - b)
Proof:
a⁴ - b⁴ = (a²)² - (b²)²
= (a² + b²)(a² - b²)
= (a² + b²)(a + b)(a - b)
This identity shows that the difference of fourth powers can be factored into the product of sum of squares and difference of squares.
3. a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)
Proof:
Let's expand the right side:
(a + b + c)(a² + b² + c² - ab - bc - ca)
= a³ + ab² + ac² - a²b - abc - a²c
+ a²b + b³ + bc² - ab² - b²c - abc
+ a²c + b²c + c³ - abc - bc² - ac²
= a³ + b³ + c³ - 3abc
This identity is particularly useful when a + b + c = 0, as it simplifies to a³ + b³ + c³ = 3abc.
Example 7
If a + b + c = 0, prove that a³ + b³ + c³ = 3abc
Solution:
Using the identity:
a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)
Since a + b + c = 0:
a³ + b³ + c³ - 3abc = 0
Therefore, a³ + b³ + c³ = 3abc
Applications of Algebraic Identities
Algebraic identities are used in various mathematical and real-world applications.
Common Applications
- Simplifying complex expressions
- Solving equations
- Factorization
- Geometric proofs
- Physics and engineering calculations
Example 8
Find the value of (1001)² using algebraic identities.
Solution:
(1001)² = (1000 + 1)²
Using (a + b)² = a² + 2ab + b²:
= 1000² + 2(1000)(1) + 1²
= 1,000,000 + 2,000 + 1
= 1,002,001
Practice Questions
Put your understanding of Algebraic Identities to the test with our comprehensive set of 20 practice questions, ranging from basic to advanced difficulty.
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