Answer: \(a=\frac12\), \(b=54\), \(c=\frac{27}{2}\).

Answer: \(a=\frac12\), \(b=54\), \(c=\frac{27}{2}\).

Use the binomial expansion:

\((3+ax)^4=3^4+4(3^3)(ax)+6(3^2)(ax)^2+4(3)(ax)^3+(ax)^4.\)

So the first four terms are

\(81+108ax+54a^2x^2+12a^3x^3.\)

Compare the coefficient of \(x^3\):

\(12a^3=\frac32.\)

Hence

\(a^3=\frac18,\qquad a=\frac12.\)

Then

\(b=108a=108\left(\frac12\right)=54,\)

and

\(c=54a^2=54\left(\frac14\right)=\frac{27}{2}.\)