Interpret the expression correctly

The given expansion is $\left(2^{1/3}+3^{1/4}\right)^{12}.$

We need the sum of all rational terms in its binomial expansion.

Write the general term

In the expansion of $(a+b)^{12}$, the general term is $T_{r+1}=\binom{12}{r}(2^{1/3})^{12-r}(3^{1/4})^r, \qquad r=0,1,2,\dots,12.$

Simplifying powers, $T_{r+1}=\binom{12}{r}2^{\frac{12-r}{3}}3^{\frac{r}{4}}.$

Condition for a term to be rational

For $T_{r+1}$ to be rational, both exponents must be integers:

• $\frac{12-r}{3} \in \mathbb{Z}$
• $\frac{r}{4} \in \mathbb{Z}$

So,

• $12-r$ must be divisible by $3$, i.e. $r \equiv 0 \pmod 3$
• $r$ must be divisible by $4$, i.e. $r \equiv 0 \pmod 4$

Therefore, $r$ must be divisible by $\operatorname{lcm}(3,4)=12$.

Since $0\le r\le 12$, possible values are $r=0,\ 12.$

Find the rational terms

• For $r=0$: $T_1=\binom{12}{0}(2^{1/3})^{12}=2^4=16.$

• For $r=12$: $T_{13}=\binom{12}{12}(3^{1/4})^{12}=3^3=27.$

Sum of rational terms

$16+27=43.$

Compare with stored answer

Our derived answer is 43, which matches option D.

Hence, the stored correct answer is correct.