JEE Math Practice Question
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 :
- A.724
- B.742
- C.693
- D.639
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