Table of Contents
Learn Algebra Formula
![]() |
Learn Algebra Formula |
Learn How algebra formulas are derived
1.(a-b) 2 = (a-b) (a-b)
= a. a- a. b- b. a+ b .b
= a2-ab-ba+b2
= a2-2ab+b2
2.(a + b) 2 = (a+ b) (a+ b)
= a. a+ a. b + b. a+ b .b
= a2+ab+ba+b2
= a2+2ab+b2
3. 3.(a+ b)2-2ab = a2+b2
{(a+ b) (a+ b)}-2ab ———left side of equal symbol
{( a. a+ a. b+ b .a + b .b)}-2ab
(a2+ab+ba+b2)-2ab
(a2+2ab+ b2)-2ab
a2+2ab+ b2-2ab —–open the bracket
a2+b2
3. 4. (a-b)(a +b) = a2– b2
(a- b)(a +b) ———left side of equal symbol
a. a+ a. b -b. a-b. b
a2+ab-ab-b2 —- +ab and –ab cancelled
a2– b2
3. 5. (a+ b+ c)2 = a2+b2+c2+2ab+2bc+2ac
( a+ b+ c) ( a+ b+ c) —- left side of equal symbol
(a .a +a .b +a .c +b. a+ b .b+ b .c+ c .a +c .b +c .c)
a2+ab+ac+ba+b2+ bc+ca+cb+c2
a2+b2+c2+2ab+2bc+2ca — ab= ba , bc = cb and ca =ac
3. 6.(a-b-c)2=a2+b2+c2-2ab+2bc-2ac
(a-b-c)2 = (a-b-c)(a-b-c) —we can write like this.
=(a. a- a. b-a. c-b. a+ b. b +b .c – c .a+ c .b + c .c
=a2-ab-ac-ba+b2+bc-ca+cb+c2
=a2+b2+c2-2ab+2bc-2ac —our formula
7. (a+ b)3= (a+ b)(a+ b)(a+ b)
= {(a+ b) (a+ b)} (a+ b)
= (a. a+ a. b+ b. a+ b. b) (a+ b)
= (a2 + ab+ba+b2) (a + b)
= (a2+2ab+b2) (a +b)
= a2.a+a2.b+2ab.a+2ab.b+b2.a+b2.b
= a3+a2b+2a2b+2ab2+b2a+b3
= a3+3a2b+3ab2+b3 ———our formula
following this we can derive all the formulas of algebra.
Algebraic Expressions:-
1. 2x means x + x, i.e. double of x. Broadly, if m is a positive whole number, then mx means m times the x. mx is also called the product of m and x.
In x m, m is called exponent and x is called base. Later, the meanings of mx and xm are expanded and are also given in those situations when m is any number of minus, different, irrational etc.
Laws of Exponents:
1.(x)m .(x)n =(x)m+ n
2. (x)m/(x)n =(x)m- n
3. (x y)m= x m y m
4. (x m)n = x m n
Basic rules and properties of Algebra.
(1) Closure: If a set has two elements x and y, then x * y is also an element of the same set.
(2) Commutativity: x * y = y * x
(3) Associativity: If x, y, z are components of a set, then (x * y) * z = x * (y * z)
4) Existence of identity: The set must have such a component that a * e = e * a = a
(5) Existence of inverse: In a set, any element corresponding to any component a has a-1 such that a * a-1 = a-1 * a = e
(6) Distribution rules against first operation and second operation: x (y * z) = (x y) * (x z) and (y* z) x = (y x) * (z x)
- (a + b)2 = a2 + 2ab + b2
- (a – b)2 = a2 – 2ab +b2
- a2– b2=(a + b)(a – b)
- (a+ b)3 = a3 + 3a2b + 3ab2 + b3
-
(a-b)3=a3-3a2b+3ab2-b3
- a3+b3=(a+ b)(a2+ab+b2)
- (a+ b+ c)2 = a2+b2+c2+2ab+2bc+2ac