The sum of all rational terms in the expansion of $\bigl(1+2^{1/3}+3^{1/2}\bigr)^{6}$ is equal to \rule{2cm}{0.4pt}.
Step-by-Step Solution
Key Concept: General term: $\dfrac{6!}{\alpha!\beta!\gamma!}\,1^{\alpha}\,(2^{1/3})^{\beta}\,(3^{1/2})^{\gamma}$ with $\alpha+\beta+\gamma=6.$ Rational $\Leftrightarrow 3\mid\beta$ and $2\mid\gamma.$
Rational iff $\beta\in\{0,3,6\}$ AND $\gamma\in\{0,2,4,6\}$, with $\alpha+\beta+\gamma=6.$ Listing $(\alpha,\beta,\gamma)$:
$(6,0,0)$: $\dfrac{6!}{6!}\cdot 1=1.$
$(4,0,2)$: $\dfrac{6!}{4!0!2!}\cdot 3=15\cdot 3=45.$
$(2,0,4)$: $\dfrac{6!}{2!0!4!}\cdot 9=15\cdot 9=135.$
$(0,0,6)$: $\dfrac{6!}{0!0!6!}\cdot 27=1\cdot 27=27.$
$(3,3,0)$: $\dfrac{6!}{3!3!0!}\cdot 2=20\cdot 2=40.$
$(1,3,2)$: $\dfrac{6!}{1!3!2!}\cdot 2\cdot 3=60\cdot 6=360.$
$(0,6,0)$: $\dfrac{6!}{0!6!0!}\cdot 4=1\cdot 4=4.$
Sum: $1+45+135+27+40+360+4=612.$
Correct Answer: 612