To find the coefficient of \(x^3\) in the expansion of \((a + x)^5\), we use the binomial theorem:

\(\text{Coeff of } x^3 \text{ in } (a + x)^5 = \binom{5}{3} a^{5-3} x^3 = 10a^2\)

For \((2 - x)^6\), the coefficient of \(x^3\) is:

\(\text{Coeff of } x^3 \text{ in } (2 - x)^6 = \binom{6}{3} (2)^{6-3} (-1)^3 x^3 = -160\)

Adding these coefficients gives the total coefficient of \(x^3\):

\(10a^2 - 160 = 90\)

Solving for \(a\):

\(10a^2 = 250\)

\(a^2 = 25\)

\(a = 5\)