If the constant term in the expansion of $$\left(\frac{\sqrt[5]{3}}{x} + \frac{2x}{\sqrt[3]{5}}\right)^{12}$$, $$x \neq 0$$, is $$\alpha \times 2^8 \times \sqrt[5]{3}$$, then $$25\alpha$$ is equal to :

We need the constant term in the expansion of $$\left(\frac{\sqrt[5]{3}}{x} + \frac{2x}{\sqrt[3]{5}}\right)^{12}$$.

Using binomial theorem, the general term is:

$$ T_{r+1} = \binom{12}{r} \left(\frac{3^{1/5}}{x}\right)^{12-r} \left(\frac{2x}{5^{1/3}}\right)^r $$

$$ T_{r+1} = \binom{12}{r} \cdot 3^{(12-r)/5} \cdot x^{-(12-r)} \cdot \frac{2^r \cdot x^r}{5^{r/3}} $$

$$ T_{r+1} = \binom{12}{r} \cdot \frac{3^{(12-r)/5} \cdot 2^r}{5^{r/3}} \cdot x^{r-(12-r)} = \binom{12}{r} \cdot \frac{3^{(12-r)/5} \cdot 2^r}{5^{r/3}} \cdot x^{2r-12} $$

For the constant term, set $$2r - 12 = 0 \Rightarrow r = 6$$. Thus

$$ T_7 = \binom{12}{6} \cdot \frac{3^{(12-6)/5} \cdot 2^6}{5^{6/3}} $$

which becomes

$$ T_7 = 924 \cdot \frac{3^{6/5} \cdot 64}{5^2} $$

and simplifies to

$$ T_7 = 924 \cdot \frac{64 \cdot 3^{6/5}}{25}. $$

Since $$3^{6/5} = 3^{1+1/5} = 3 \cdot 3^{1/5} = 3 \cdot \sqrt[5]{3},$$ we get

$$ T_7 = 924 \cdot \frac{64 \cdot 3 \cdot \sqrt[5]{3}}{25} = 924 \cdot \frac{192 \cdot \sqrt[5]{3}}{25} = \frac{924 \times 192}{25} \cdot \sqrt[5]{3}. $$

Computing $$924 \times 192 = 924 \times 200 - 924 \times 8 = 184800 - 7392 = 177408$$ gives

$$ T_7 = \frac{177408}{25} \cdot \sqrt[5]{3}. $$

This equals $$\alpha \times 2^8 \times \sqrt[5]{3} = 256\alpha \cdot \sqrt[5]{3},$$ so

$$256\alpha = \frac{177408}{25}$$

and

$$\alpha = \frac{177408}{25 \times 256} = \frac{177408}{6400} = 27.72.$$

Multiplying by 25 yields

$$25\alpha = \frac{177408}{256} = 693.$$

Verifying,

$$693 \times 256 = 693 \times 250 + 693 \times 6 = 173250 + 4158 = 177408.$$

Correct!

The correct answer is Option (4): 693.