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

Answer: (C)

693

Answer 2

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

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

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