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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