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

Expansion of $$\left(x^{2/3} + \frac{\alpha}{x^3}\right)^{22}$$. The term independent of $$x$$ equals 7315. Find $$|\alpha|$$.

The general term of the binomial expansion is $$T_{r+1} = \binom{22}{r} \left(x^{2/3}\right)^{22-r} \left(\frac{\alpha}{x^3}\right)^r = \binom{22}{r} \alpha^r \cdot x^{\frac{2(22-r)}{3} - 3r}$$.

Setting the exponent of $$x$$ to zero gives $$\frac{2(22-r)}{3} - 3r = 0$$, so $$\frac{44 - 2r}{3} = 3r$$, $$44 - 2r = 9r$$, $$44 = 11r$$ and $$r = 4$$.

For $$r = 4$$ the term is $$T_5 = \binom{22}{4} \alpha^4 = 7315$$ and since $$\binom{22}{4} = \frac{22 \times 21 \times 20 \times 19}{4!} = \frac{175560}{24} = 7315$$, we have $$7315 \cdot \alpha^4 = 7315$$, yielding $$\alpha^4 = 1$$ and hence $$|\alpha| = 1$$.

The answer is 1.