The general term in the binomial expansion of \((1 + ax)^6\) is given by:

\(T_k = \binom{6}{k} (1)^{6-k} (ax)^k = \binom{6}{k} a^k x^k\)

For the term in \(x\), we have \(k = 1\):

\(T_1 = \binom{6}{1} a^1 x^1 = 6ax\)

We know the coefficient of \(x\) is \(-30\), so:

\(6a = -30\)

Solving for \(a\):

\(a = -5\)

Now, find the coefficient of \(x^3\) (\(k = 3\)):

\(T_3 = \binom{6}{3} a^3 x^3\)

\(\binom{6}{3} = \frac{6 \times 5 \times 4}{3!} = 20\)

Substitute \(a = -5\):

\(T_3 = 20(-5)^3 x^3\)

\(T_3 = 20(-125)x^3\)

\(T_3 = -2500x^3\)

Thus, the coefficient of \(x^3\) is \(-2500\).