The formula for abbreviated multiplication
Formulas of abbreviated multiplication allow you to perform calculations much faster.
The most commonly used abbreviated multiplication formulas:
(a + b)^{2 }= a^{2 }+ 2from + b^{2}
(a − b)^{2 }= a^{2 }− 2from + b^{2}
(a+b+c)^{2 }= a^{2 }+ b^{2 }+ c^{2 }+ 2from + 2ac + 2bc
a^{2 }− b^{2 }= (a + b)(a − b)
(a + b)^{3 }= a^{3 }+ 3a^{2}b + 3from^{2 }+ b^{3}
(a − b)^{3 }= a^{3 }− 3a^{2}b + 3from^{2 }− b^{3}
a^{3 }+ b^{3 }= (a + b)(a^{2 }−from + b^{2})
a^{3 }− b^{3 }= (a − b)(a^{2 }+ from + b^{2})
The abbreviated multiplication formulas are helpful for multiplying or expanding algebraic expressions. They facilitate efficient counting. There are a lot of these patterns. We list a few below, which are used most often.
The square of the sum of the numbers

(a + b)^{2 }= a^{2 }+ 2from + b^{2 }e.g: 31^{2 }= (30+1)^{2 }= 30^{2}+2×30+1 = 900+60+1 = 961

does not occur equality: (a+b)^{2} = a^{2} + b^{2 }e.g 25 = (3+2)^{2} ≠ 3^{2} + 2^{2 }= 13

justification of the formula by the bill:
(a + b)^{2 }= (a + b) × (a + b)^{ }= aa^{ }+ from + ba + bb = a^{2 }+ 2from + b^{2}
The square of the difference of numbers

(a – b)^{2 }= a^{2 }– 2from + b^{2 }e.g: 29^{2 }= (301)^{2 }= 30^{2}2×30+1 = 90060+1 = 841

does not occur equality: (ab)^{2} = a^{2} – b^{2 }e.g 1 = (32)^{2} ≠ 3^{2} – 2^{2 }= 5

justification of the formula:
(a – b)^{2 }= (a – b) × (a – b)^{ }= aa^{ }– from – ba + bb = a^{2 }– 2from + b^{2}
A square of the sum of three numbers

(a+b+c)^{2 }= a^{2 }+ b^{2 }+ c^{2 }+ 2from + 2ac + 2bc
e.g: 111^{2 }= (100+10+1)^{2 }= 100^{2 }+ 10^{2 }+1 +2×100×10 + 2×100 + 2×10 = 10000 + 100 + 1 + 2000 + 200 + 20 = 12321 
does not occur equality: (a+b+c)^{2} = a^{2} + b^{2 }+ c^{2 }e.g 36 = (3+2+1)^{2} ≠ 3^{2} + 2^{2 }+ 1^{2} = 14

justification of the formula:
(a + b + c)^{2 }= (a + b + c) × (a + b + c)^{ }= aa^{ }+ from + ac + ba + bb + bc + that + cb + cc = a^{2 }+ b^{2 }+ c^{2 }+ 2from + 2ac + 2bc
Product of the sum and difference of numbers = Difference of squares of numbers

(a + b)×(a – b) = a^{2 }– b^{2 }e.g: 101×99 = (100+1)×(1001) = 100^{2 }– 1 = 9999

justification of the formula :
(a + b) × (a – b)^{ }= aa^{ }– from + ba – bb = a^{2 }– b^{2}
A cube of the sum of numbers

(a + b)^{3 }= a^{3 }+ 3a^{2}b + 3from^{2 }+ b^{3 }e.g: 101^{3 }= (100+1)^{3} =^{ }100^{3 }+ 3×100^{2 }+ 3×100 + 1 =
= 1000000 + 30000 + 300 + 1 = 1030301 
does not occur equality: (a+b)^{3} = a^{3} + b^{3 }e.g 125 = (3+2)^{3} ≠ 3^{3} + 2^{3 }= 35

justification of the formula by the bill:
(a + b)^{3 }= (a + b) × (a + b) × (a + b)^{ }= (aa^{ }+ from + ba + bb) × (a + b) = aaa^{ }+ aab + aba + fig + baa^{ }+ chapter + bba + bbb =
= a^{3 }+ 3a^{2}b + 3from^{2 }+ b^{3}
Cube of number difference
 (a – b)^{3 }= a^{3 }– 3a^{2}b + 3from^{2 }– b^{3}^{
}e.g: 99^{3} = (1001)^{3} = 100^{3 }– 3×100^{2 }+ 3×100 – 1 =
= 1000000 – 30000 + 300 – 1 = 970299
Sum of cubes of numbers
a^{3 }+ b^{3 }= (a + b)×(a^{2 }– from + b^{2})
justification of the formula:
(a + b)×(a^{2 }– from + b^{2}) = aa^{2 }– aab + from^{2 }+ ba^{2 }– chapter + bb^{2}= a^{3} – a^{2}b + from^{2 }+ a^{2}b – from^{2 }+ b^{3} =
= a^{3} + b^{3}
The difference of the cubes of numbers
a^{3 }– b^{3 }= (a – b)×(a^{2 }+ from + b^{2})
justification of the formula:
(a – b)×(a^{2 }+ from + b^{2}) = aa^{2 }+ aab + from^{2 }– ba^{2 }– chapter – bb^{2} = a^{3} + a^{2}b + from^{2 }– a^{2}b – from^{2 }– b^{3} =
= a^{3} – b^{3}
Difference of fourth powers of numbers
a^{4 }– b^{4 }= (a – b)×(a^{3 }+ a^{2}b + from^{2} + b^{3}) = (a + b)×(a^{3} – a^{2}b + from^{2} – b^{3})
Sum nthese powers of numbers (for n odd!!!)
a^{n} + b^{n} = (a + b) (a^{n}^{1} – a^{n}^{2}b + a^{n}^{3}b^{2} – … + b^{n}^{1})
Difference nthese powers of numbers (for n even!!!)
a^{n} – b^{n} = (a + b) (a^{n}^{1} – a^{n}^{2}b + a^{n}^{3}b^{2} – … + b^{n}^{1})
Difference nthese powers of numbers (for everyone n natural)
a^{n} – b^{n} = (a – b) (a^{n}^{1} + a^{n}^{2}b + a^{n}^{3}b^{2} + … + a^{2}b^{n}^{3} + from^{n}^{2} + b^{n}^{1})