Answer :

We know that-

Hence

= a^{4} + 4a^{3}b + 6a^{2}b^{2} + 4ab^{3} + b^{4}

Thus, (a + b)^{4} = a^{4} + 4a^{3}b + 6a^{2}b^{2} + 4ab^{3} + b^{4} …(1)

Putting , we get-

…(2)

Now Solving separately

From (1)

(a + b)^{4} = a^{4} + 4a^{3}b + 6a^{2}b^{2} + 4ab^{3} + b^{4}

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

we know that-

(a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}

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

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

Thus,

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