If the coefficients of the middle terms in the binomial expansions of $$(1 + \alpha x)^{26}$$ and $$(1 - \alpha x)^{28}$$, $$\alpha \neq 0$$, are equal, then the value of $$\alpha$$ is:

For $$(1 + \alpha x)^{26}$$, the middle term index is $$r = \frac{26}{2} = 13$$.

The middle term is $$T_{14} = \binom{26}{13} (\alpha x)^{13}$$

$$C_1 = \binom{26}{13} \alpha^{13}$$

Similarly, $$C_2 = \binom{28}{14} \alpha^{14}$$

$$\binom{26}{13} \alpha^{13} = \binom{28}{14} \alpha^{14}$$

$$\alpha = \frac{\binom{26}{13}}{\binom{28}{14}}$$

$$\alpha = \frac{\frac{26!}{13! \, 13!}}{\frac{28!}{14! \, 14!}}$$

$$\alpha = \frac{26!}{28!} \times \frac{14! \times 14!}{13! \times 13!}$$

$$\alpha = \frac{1}{28 \times 27} \times (14 \times 14) = \frac{7}{27}$$