maths

The formula for abbreviated multiplication

The formula for abbreviated multiplication

The most important abbreviated multiplication formulas.

Formulas of abbreviated multiplication allow you to perform calculations much faster.
The most commonly used abbreviated multiplication formulas:

(a + b)2 = a2 + 2ab + b2

(a − b)2 = a2 − 2ab + b2

(a+b+c)2 = a2 + b2 + c2 + 2ab + 2Ac + 2Bc

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)

a3 − b3 = (a − b)(a2 + ab + b2)

 

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 = a2 + 2ab + b2
    e.g: 312 = (30+1)2 = 302+2×30+1 = 900+60+1 = 961
  • does not occur equality: (a+b)2 = a2 + b2
    e.g 25 = (3+2)2 ≠ 32 + 22 = 13
  • justification of the formula by the bill:
    (a + b)2 = (a + b) × (a + b) =     Aa + ab + Ba + Bb = a2 + 2ab + b2

The square of the difference of the numbers

  • (a – b)2 = a2 – 2ab + b2
    e.g: 292 = (30-1)2 = 302-2×30+1 = 900-60+1 = 841
  • does not occur equality: (a-b)2 = a2 – b2
    e.g 1 = (3-2)2 ≠ 32 – 22 = 5
  • justification of the formula:
    (a – b)2 = (a – b) × (a – b) = Aa – ab – Ba + Bb = a2 – 2ab + b2

     

A square of the sum of three numbers

  • (a+b+c)2 = a2 + b2 + c2 + 2ab + 2Ac + 2Bc
    e.g: 1112 = (100+10+1)2 = 1002 + 102 +1 +2×100×10 + 2×100 + 2×10 = 10000 + 100 + 1 + 2000 + 200 + 20 = 12321
  • does not occur equality: (A+B+c)2 = a2 + b2 + c2
    e.g 36 = (3+2+1)2 ≠ 32 + 22 + 12 = 14
  • justification of the formula:
    (a + b + c)2 = (a + b + c) × (a + b + c) = Aa + ab + Ac + Ba + Bb + Bc + ca + Cb + cc = a2 + b2 + c2 + 2ab + 2Ac + 2Bc

Product of the sum and difference of numbers = Difference of squares of numbers

  • (a + b)×(a – b) = a2 – b2
    e.g: 101×99 = (100+1)×(100-1) = 1002 – 1 = 9999
  • justification of the formula :
    (a + b) × (a – b) = Aa – ab + Ba – Bb = a2 – b2

A cube of the sum of numbers

  • (a + b)3 = a3 + 3a2b + 3ab2 + b3
    e.g: 1013 = (100+1)3 = 1003 + 3×1002 + 3×100 + 1 =
    = 1000000 + 30000 + 300 + 1 = 1030301
  • does not occur equality: (a+b)3 = a3 + b3
    e.g 125 = (3+2)3 ≠ 33 + 23 = 35
  • justification of the formula by the bill:
    (a + b)3 = (a + b) × (a + b) × (a + b)     = (Aa + ab + Ba + Bb) × (a + b) = Aaa + aab + aba + fig + Baa + Bab + Bba + bbb =
    = a3 + 3a2b + 3ab2 + b3

Cube of number difference

  • (a – b)3 = a3 – 3a2b + 3ab2 – b3
    e.g: 993 = (100-1)3 = 1003 – 3×1002 + 3×100 – 1 =
    = 1000000 – 30000 + 300 – 1 = 970299

Sum of cubes of numbers

a3 + b3 = (a + b)×(a2 – ab + b2)

justification of the formula:

(a + b)×(a2 – ab + b2) = Aa2 – aab + ab2 + Ba2 – Bab + Bb2= a3 – a2b + ab2 + a2b – ab2 + b3 =
= a3 + b3

The difference of the cubes of numbers

a3 – b3 = (a – b)×(a2 + ab + b2)

justification of the formula:

(a – b)×(a2 + ab + b2) = Aa2 + aab + ab2 – Ba2 – Bab – Bb2 = a3 + a2b + ab2 – a2b – ab2 – b3 =
= a3 – b3

Difference of fourth powers of numbers

a4 – b4 = (a – b)×(a3 + a2b + ab2 + b3) = (a + b)×(a3 – a2b + ab2 – b3)

 

Addition n-these powers of numbers (For n odd!!!)

an + bn = (a + b) (an-1 – an-2b + an-3b2 – … + bn-1)

 

Difference n-these powers of numbers (For n even!!!)

an – bn = (a + b) (an-1 – an-2b + an-3b2 – … + bn-1)

 

Difference n-these powers of numbers (for everyone n natural)

an – bn = (a – b) (an-1 + an-2b + an-3b2 + … + a2bn-3 + abn-2 + bn-1)