Question

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

• A. 639
• B. 724
• C. 693
• D. 742

Answer 1

The Correct Option is C

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

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

Given \( r = 6 \), \[ T_7 = \binom{12}{6} \frac{(3^{6/5})(2^6)(x^6)}{x^6 (5^2)} = \binom{12}{6} \frac{3^6 \cdot 2^6}{5^2} \]

Simplifying further, \[ T_7 = \frac{9 \cdot 11 \cdot 7}{25} \cdot 2^8 \cdot 3^{1/5} \]

Finally, equating, \[ 25\alpha = 693 \]

Final Answer: 693