Algebraic Identites

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Formulae for binomial

   1.       (a + b)2 = a2  + 2ab + b2

Proof :    Taking L.H.S
             (a + b)2= (a + b) . (a + b)
                          = a (a + b ) + b (a + b)
                          = a2 + ab + ab + b2
           (a + b)2  = a2  + 2ab + b2      …….. (i)

   2.       (a – b)2 = a2 – 2ab + b2

Proof :  Putting b = – b in equn  (i)
        (a – b)2  = a2  + 2a(–b) + (– b)2
        (a – b)2 = a2  –  2ab + b2

   3.       a2 – b2 = (a + b) . (a – b)

Proof :  Taking R.H.S
       (a + b) . (a – b) = a (a – b) + b (a – b)
                                 = a2 –  ab + ab – b2
                                 = a2 – b2

   4.       (a + b)3 = a3 + b3 + 3ab (a + b) = a3 + b3 + 3a2b + 3ab2

Proof : Taking L.H.S
   (a + b)3= (a + b) . (a + b)2
                = (a + b) . (a2  + 2ab + b2)
                = a (a2  + 2ab + b2) + b (a2  + 2ab + b2)
                = a3 + 2a2b + ab2 + a2b + 2ab2 + b3
                   = a3 + 3a2b + 3ab2 + b3
   (a + b)3= a3 + b3 + 3a2b + 3ab2 = a3 + b3 + 3ab (a + b)..(ii)

   5.        (a – b)3 = a3 – b3 – 3ab (a – b) = a3 – b3 – 3a2b + 3ab2

Proof : Putting b = – b in equation (ii)
    (a – b)3 = a3 +(– b)3 + 3a(–b) [a +(– b)]
    (a – b)3 = a3 – b3 – 3ab (a – b) = a3 – b3 – 3a2b + 3ab2.....(iii)

   6.       (a3  + b3) = (a + b)(a2 – ab + b2)

Proof : From (ii)
                     (a + b)3 = a3 + b3 + 3ab (a + b)
(a + b)3 – 3ab (a + b)= a3 + b3
                      a3 + b3 = (a + b)3 – 3ab (a + b)
                                   = (a + b) [(a + b)2 – 3ab]
                                   = (a + b) [ a2 + 2ab + b2 – 3ab]
                       (a3 + b3) = (a + b) ( a2 – ab + b2)

   7.       (a3  – b3) = (a – b)(a2 + ab + b2)

Proof : From (iii)
             (a – b)3 = a3 – b3 – 3ab (a – b)
(a – b)3 + 3ab (a – b)= a3 – b3
a3 – b3 = (a – b)3 + 3ab (a – b)
             = (a – b) [(a – b)2 + 3ab]
             = (a – b) [ a2 – 2ab + b2 + 3ab]
     (a3 – b3) = (a – b) ( a2 + ab + b2)

formula for Trinomial

     8.       (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

Proof : Taking L.H.S
        (a + b + c)2 = (a + b)2 + c2 + 2 (a + b) . c
                           = a2 + 2ab + b2 + c2 + 2ac + 2bc
        (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

    9.    (a3 + b3 + c3 – 3abc) = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

Proof: Taking L.H.S
 (a3 + b3+ c3 – 3abc) = (a + b)3 – 3ab (a + b) + c3 – 3abc
                      = (x3 + c3) – 3ab (x + c), where (a + b)= x
                      = (x + c)(x2 – xc + c2) – 3ab (x + c)
                      = (x + c)[(x2 – xc + c2) – 3ab]
                      = (a + b + c)[(a + b)2 – (a + b)c + c2 – 3ab]
                      = (a + b + c)[a2 + 2ab + b2 – ac – bc + c2 – 3ab] 
 (a3 + b3+ c3 – 3abc) = ( a + b + c)(a2 + b2 + c2 – ab – bc – ca)

   10.   If (a + b + c) = 0 ⇒ a3 + b3 + c3 = 3abc

Proof :  We have
                             (a + b + c) = 0
                                    (a + b) = – c
Cubing both sides
                                  (a + b)3 = (­– c)3
            a3 + b3 + 3ab (a + b) = – c3
                a3 + b3 + 3ab (­– c) =  – c3
                       a3 + b3 – 3abc = – c3

                          a3 + b3 + c3 = 3abc

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