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2014-10-01T21:54:59+05:30

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(a+b+c)²
=(a+b+c) × (a+b+c)
=[a×(a+b+c)] + [b×(a+b+c)] +[c×(a+b+c)] (by distributive law)
= [a² + ab + ac] + [ ba +b² + bc] + [ ca + cb + c²] 
= a² + ab + ac + ba +b² + bc + ca + cb + c²
= a² + ab + ac + ab +b² + bc + ac + bc + c²   (ab=ba)
= a² + b² + c² + ab + ab + bc + bc + ac + ac
= a² + b² + c² + 2ab + 2bc + 2ca
2 5 2
ok now?
its enough if u can add more its better
thank u i will surely mark as best
:P
thank alot thank u
2014-10-02T07:45:54+05:30
(a+b+c)²

let a+b=x  ,c=y

substitute the values in the equation

(x+y)²=x²+y²+2xy

(a+b)²+c²+2c(a+b)

a²+b²+2ab+c²+2ac+2bc

a²+b²+c²+2ab+2bc+2ca


1 5 1