The coefficient of $$x^{18}$$ in the expansion of $$\left(x^4 - \dfrac{1}{x^3}\right)^{15}$$ is ______.

The general term of the expansion $$\left(x^4 - \frac{1}{x^3}\right)^{15}$$ is:

$$ T_{r+1} = \binom{15}{r}(x^4)^{15-r}\left(-\frac{1}{x^3}\right)^r = \binom{15}{r}(-1)^r x^{4(15-r)-3r} = \binom{15}{r}(-1)^r x^{60-7r} $$

For the coefficient of $$x^{18}$$:

$$ 60 - 7r = 18 \implies 7r = 42 \implies r = 6 $$

The coefficient is:

$$ \binom{15}{6}(-1)^6 = \binom{15}{6} = \frac{15!}{6! \cdot 9!} = 5005 $$

The coefficient of $$x^{18}$$ is 5005.