JEE Main 20245 Apr, Evening shiftMathematicsMediumBinomial Theorem and Its Simple Applications131× since 2002If the constant term in the expansion of (35x+2x53)12,x≠0\left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[3]{5}}\right)^{12}, x \neq 0(x53+352x)12,x=0, is α×28×35\alpha \times 2^8 \times \sqrt[5]{3}α×28×53, then 25α25 \alpha25α is equal to :AOption A. 724BOption B. 742COption C. 693DOption D. 639CheckSkip · show solution
