Sun. Sep 8th, 2024

Dec 7, 2017

Basic Algebraic Formulas

Basic Algebraic Formulas

1. (a + b)² = a² + 2ab + b²
= (a – b)² + 4ab
2. (a – b)² = a² + 2ab + b²
= (a – b)² + 4ab
3. (a + b)³ = a³ + 3a²b + 3ab² + b³
= a³ + b³ + 3ab(a + b)
4. (a – b)³  = a³ – 3a²b + 3ab² – b³
= a³ – b³ – 3ab(a – b)
5. a² + b²  = (a + b)² – 2ab
= (a – b)² + 2ab
= $\frac{1}{2}$ [ (a + b)² + (a – b)² ]
6. a² – b²   = (a + b)(a – b)
7. a³ + b³  = (a + b)(a² – ab + b²)
= (a + b)³ – 3ab(a + b)
8. a³ – b³   = (a – b)(a² + ab + b²)
= (a – b)³ + 3ab(a – b)
9. 4ab       = (a + b)² – (a – b)²
10. ab        = [(a + b)/2]² – [(a – b)/2]²

Some Special Formulas:

1. (a + b + c)2

= a² + b² + c² + 2ab + 2bc + 2ca
2. a² + b² + c² + ab + bc + c

= $\frac{1}{2}$ [(a + b)² + (b + c)² + (c + a)²] 3. a² + b² + c² – ab – bc – ca

=$\frac{1}{2}$ [(a – b)² + (b – c)² + (c – a)²] 4. a³ + b³ + c³ – 3abc

= (a + b + c)(a² + b² + c² – ab – bc – ca)
= $\frac{1}{2}$ (a + b + c)[(a – b)² + (b – c)² + (c – a)²]

CIT-002 Solved Assignment 2024-25

Insert math as
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