Q. 173.8( 11 Votes )

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Answer :

We know that-



Hence






= a4 + 4a3b + 6a2b2 + 4ab3 + b4


Thus, (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 …(1)


Putting , we get-



…(2)


Now Solving separately


From (1)


(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4


putting a = 1 & b = (x/2), we get-





we know that-


(a + b)3 = a3 + 3a2b + 3ab2 + b3


putting a = 1 & b = (x/2), we get-




Substituting the value of in (2), we get-







Thus,

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