1. Write the general term

For $\left(\frac{\sqrt[5]{3}}{x}+\frac{2x}{\sqrt[3]{5}}\right)^{12},$ the general term is $T_{r+1}=\binom{12}{r}\left(\frac{\sqrt[5]{3}}{x}\right)^{12-r}\left(\frac{2x}{\sqrt[3]{5}}\right)^r.$

2. Simplify the power of $x$

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

So the power of $x$ is $x^{-12+r+r}=x^{2r-12}.$

For the constant term, exponent of $x$ must be zero: $2r-12=0 \implies r=6.$

3. Find the constant term

Substitute $r=6$: $T_7=\binom{12}{6}\left(\frac{\sqrt[5]{3}}{x}\right)^6\left(\frac{2x}{\sqrt[3]{5}}\right)^6.$

Now, $\binom{12}{6}=924,$ $(\sqrt[5]{3})^6=3^{6/5}=3\cdot 3^{1/5}=3\sqrt[5]{3},$ $\left(\frac{1}{\sqrt[3]{5}}\right)^6=5^{-2}=\frac{1}{25},$ $2^6=64.$

Hence, $\text{constant term}=924\cdot 64\cdot \frac{3\sqrt[5]{3}}{25}.$

So, $\text{constant term}=\frac{924\cdot 3\cdot 2^6}{25}\sqrt[5]{3}.$

Since the question says this equals $\alpha\times 2^8\times \sqrt[5]{3},$ we equate: $\alpha\cdot 2^8=\frac{924\cdot 3\cdot 2^6}{25}.$

Divide both sides by $2^6$: $\alpha\cdot 2^2=\frac{924\cdot 3}{25}$ $4\alpha=\frac{2772}{25}$ $\alpha=\frac{693}{25}.$

Therefore, $25\alpha=693.$

4. Check options

• A: $724$
• B: $742$
• C: $693$
• D: $639$

So the correct option is C.