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 :
Hello 👋 Welcome to Competishun – India’s most trusted platform for JEE & NEET preparation. Need help with JEE / NEET courses, fees, batches, test series or free study material? Chat with us now 👇