If the coefficients of $$x$$ and $$x^2$$ in the expansion of $$(1 + x)^p(1 - x)^q$$, $$p, q \le 15$$, are $$-3$$ and $$-5$$ respectively, then the coefficient of $$x^3$$ is equal to ______.

We need to find the coefficient of $$x^3$$ in $$(1+x)^p(1-x)^q$$, given that the coefficients of $$x$$ and $$x^2$$ are $$-3$$ and $$-5$$ respectively.

Since the coefficient of $$x$$ is $$-3$$, $$[x^1]: \binom{p}{1}(1) + \binom{q}{1}(-1) = p - q = -3 \quad \cdots (1)$$

Next, since the coefficient of $$x^2$$ is $$-5$$, $$[x^2]: \binom{p}{2} - pq + \binom{q}{2} = \dfrac{p(p-1)}{2} - pq + \dfrac{q(q-1)}{2} = -5$$

$$p^2 - p + q^2 - q - 2pq = -10$$

$$(p - q)^2 - (p + q) = -10$$

$$9 - (p + q) = -10$$

$$p + q = 19 \quad \cdots (2)$$

From (1) and (2), $$p = 8$$, $$q = 11$$. Check: $$p, q \le 15$$. $$\checkmark$$

Next, $$[x^3] = \sum_{j=0}^{3} \binom{p}{3-j}\binom{q}{j}(-1)^j$$

$$= \binom{8}{3} - \binom{8}{2}\binom{11}{1} + \binom{8}{1}\binom{11}{2} - \binom{11}{3}$$

$$= 56 - 28 \times 11 + 8 \times 55 - 165$$

$$= 56 - 308 + 440 - 165$$

$$= 23$$

The correct answer is $$\boxed{23}$$.