If the term without $$x$$ in the expansion of $$\left(x^{\frac{2}{3}} + \frac{\alpha}{x^3}\right)^{22}$$ is $$7315$$, then $$|\alpha|$$ is equal to ______.

We consider the expansion of $$\left(x^{2/3} + \frac{\alpha}{x^3}\right)^{22}$$ and note that its general term is $$T_{r+1} = \binom{22}{r} \left(x^{2/3}\right)^{22-r} \left(\frac{\alpha}{x^3}\right)^r = \binom{22}{r} \alpha^r \cdot x^{\frac{2(22-r)}{3} - 3r}$$.

The exponent of $$x$$ in this term is $$\frac{44 - 2r}{3} - 3r = \frac{44 - 11r}{3}$$. Setting this equal to zero yields $$44 - 11r = 0 \Rightarrow r = 4$$, so the term independent of $$x$$ is $$T_5 = \binom{22}{4} \alpha^4$$.

Computing $$\binom{22}{4}$$ gives $$\binom{22}{4} = \frac{22 \times 21 \times 20 \times 19}{4!} = \frac{175560}{24} = 7315$$, hence the constant term is $$7315 \alpha^4$$. Since this equals 7315, we have $$7315 \cdot \alpha^4 = 7315 \Rightarrow \alpha^4 = 1 \Rightarrow |\alpha| = 1$$.