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

Find constant term in $\left(\dfrac{\sqrt[5]{3}}{x} + \dfrac{2x}{\sqrt[3]{5}}\right)^{12}$

General Term:

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

$= \binom{12}{r} \cdot 3^{\frac{12-r}{5}} \cdot 2^r \cdot 5^{-r/3} \cdot x^{2r-12}$

For constant term:

$2r - 12 = 0 \Rightarrow r = 6$

Put $r = 6$:

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

$= \binom{12}{6} \cdot 3 \cdot 3^{\frac{1}{5}} \cdot 2^6 \cdot 5^{-2}$

$= \binom{12}{6} \cdot \dfrac{3 \times 64}{25} \cdot \sqrt[5]{3}$

$= \alpha \times 2^8 \times \sqrt[5]{3}$

Cancel $\sqrt[5]{3}$ both sides:

$\binom{12}{6} \cdot \dfrac{3 \times 64}{25} = \alpha \times 256$

$924 \times \dfrac{192}{25} = \alpha \times 256$

$\alpha = \dfrac{924 \times 192}{25 \times 256} = \dfrac{924 \times 3}{25 \times 4} = \dfrac{2772}{100} = \dfrac{693}{25}$

$25\alpha = \boxed{693}$