Algebraic formulas
- a2 – b2 = (a – b)(a + b)
- (a+b)2 = a2 + 2ab + b2
- a2 + b2 = (a – b)2 + 2ab
- (a – b)2 = a2 – 2ab + b2
- (a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc
- (a – b – c)2 = a2 + b2 + c2 – 2ab – 2ac + 2bc
- (a + b)3 = a3 + 3a2b + 3ab2 + b3 ; (a + b)3 = a3 + b3 + 3ab(a + b)
- (a – b)3 = a3 – 3a2b + 3ab2 – b3
- a3 – b3 = (a – b)(a2 + ab + b2)
- a3 + b3 = (a + b)(a2 – ab + b2)
- (a + b)3 = a3 + 3a2b + 3ab2 + b3
- (a – b)3 = a3 – 3a2b + 3ab2 – b3
- (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4)
- (a – b)4 = a4 – 4a3b + 6a2b2 – 4ab3 + b4)
- a4 – b4 = (a – b)(a + b)(a2 + b2)
- a5 – b5 = (a – b)(a4 + a3b + a2b2 + ab3 + b4)
- If n is a natural number, an – bn = (a – b)(an-1 + an-2b+…+ bn-2a + bn-1)
- If n is even (n = 2k), an + bn = (a + b)(an-1 – an-2b +…+ bn-2a – bn-1)
- If n is odd (n = 2k + 1), an + bn = (a + b)(an-1 – an-2b +…- bn-2a + bn-1)
- (a + b + c + …)2 = a2 + b2 + c2 + … + 2(ab + ac + bc + ….
- Laws of Exponents
(am)(an) = am+n
(ab)m = ambm
(am)n = amn
- Fractional Exponents
a0 = 1
aman=am−naman=am−n
amam = 1a−m1a−m
a−ma−m = 1am1am
SOLVED EXAMPLES
Question 1: Find out the value of 52 – 32
Solution:
Using the formula a2 – b2 = (a – b)(a + b)
where a = 5 and b = 3
(a – b)(a + b)
= (5 – 3)(5 + 3)
= 2 ×× 8
= 16
Question 2: 43 ×× 42 = ?
Solution:
Using the exponential formula (am)(an) = am+n
where a = 4
43 ×× 42
= 43+2
= 45
= 1024