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HCF and LCM

Hcf

H.C.F


Q. Find H.C.F of: a4+a2b2+b4, a3-b3,  and a3+a2b +ab2


Soln:

1St exp:     a4+a2b2+b4

                      = (a2)2 + ( b2 )2 + a2b2

                      = (a2 + b2 )2 - 2a2b2 + a2 b2

                = (a2 + b2)2 - ab2

               = (a2+ ab +b2) (a2- ab + b2

2Nd exp:  a3 - b 3

              = a - b (a2+ ab + b2)

3rd exp:   a3+ a2b + ab2 

              = a (a2+ ab + b2

Therefore, H.C.F = (a2+ ab + b2 )