JEE Main 2025 — 23 January, Morning Shift — Question 22

$\text{We\; expand }(1+2^{1/3}+3^{1/2})^6=\sum_{a+b+c=6}\frac{6!}{a!\,b!\,c!}\,1^a(2^{1/3})^b(3^{1/2})^c.$

$\text{A \;term\; is\; rational\; iff\; }b/3\in\mathbb{Z}\; \text{ and\; }c/2\in\mathbb{Z},\text{\; so\; }b\in\{0,3,6\},\ c\in\{0,2,4,6\}.$

$\text{Case\; }b=0:\; (a,c)=(6,0),(4,2),(2,4),(0,6).$ $1,\; \dfrac{6!}{4!0!2!}3^1=15\cdot3=45,\; \dfrac{6!}{2!0!4!}3^2=15\cdot9=135,\;1\cdot3^3=27.$

$\text{Sum}_{b=0}=1+45+135+27=208.$

$\text{Case\; }b=3:\; (a,c)=(3,0),(1,2).$ $\dfrac{6!}{3!3!0!}2^{1}=20\cdot2=40,\; \dfrac{6!}{1!3!2!}2^{1}3^{1}=60\cdot6=360.$

$\text{Sum}_{b=3}=40+360=400.$

$\text{Case\; }b=6:\; (a,c)=(0,0):\; \dfrac{6!}{0!6!0!}2^{2}=4.$

$\text{Total\; sum\; of\; rational\; terms}=208+400+4=612.$

$\boxed{612}$