Verified Answer :

The coefficients are given by \(\dbinom 3k=\dfrac{3!}{k!(3-k)!}\) where \(k\in\{0,1,2,3\}\). These are, in order, \(\{1,3,3,1\}\).

So the expansion is

\((a+b)^3=1a^3+3a^2b+3ab^2+1b^3\)

Answer:

Required coefficient :coefficient of 1st, 2nd, 3rd and 4th term are 1, 3,3,1 respectively.

Explanation:  

Formula for binomial expansion is \((a+b)^n=^nC_0\cdot a^n\cdot b^0+^nC_1\cdota^(n-1)\cdot b^1+------b^n\)

We will substitute the values a=a, b=b and n=3  in the formula for binomial expansion we will get

\(^3C_0\cdot a^3\cdot b^0+^3C_1\cdot a^2\cdot b^1+^3C_2\cdot a^1\cdot b^2+^3C_3\cdot a^0\cdot b^3\)

After simplification of the terms we will get

\(a^3+3\cdot a^2\cdot b+3\cdot a\cdot b^2+b^3\)

Since, \(^nC_r=\frac{n!}{(r!)(n-r)!}\)

Therefore the required coefficient : coefficients of 1st, 2nd, 3rd and 4th term are 1, 3,3,1 respectively.