The sum of all rational terms in the expansion of $(1+2^{1/3}+3^{1/2})^{6}$ is equal to
Solution:
The general term in the expansion of $(1+2^{1/3}+3^{1/2})^{6}$ is given by the multinomial theorem:
$$T_{r_1, r_2, r_3} = \frac{6!}{r_1! r_2! r_3!} (1)^{r_1} (2^{1/3})^{r_2} (3^{1/2})^{r_3}$$
where $r_1 + r_2 + r_3 = 6$ and $r_1, r_2, r_3 \in \{0, 1, …, 6\}$.
For the term to be rational, the powers of 2 and 3 must be integers.
Thus, $r_2$ must be a multiple of 3 ($r_2 \in \{0, 3, 6\}$) and $r_3$ must be a multiple of 2 ($r_3 \in \{0, 2, 4, 6\}$).
We verify the possible cases for $(r_2, r_3)$:
$\Rightarrow$ Case 1: if $r_2 = 0$
Possible $r_3 \in \{0, 2, 4, 6\}$ (since $r_1 \ge 0$).
$(6, 0, 0): \frac{6!}{6!0!0!} (1)(1)(1) = 1$
$(4, 0, 2): \frac{6!}{4!0!2!} (1)(1)(3^1) = 15 \times 3 = 45$
$(2, 0, 4): \frac{6!}{2!0!4!} (1)(1)(3^2) = 15 \times 9 = 135$
$(0, 0, 6): \frac{6!}{0!0!6!} (1)(1)(3^3) = 1 \times 27 = 27$
$\Rightarrow$ Case 2: if $r_2 = 3$
Possible $r_3 \in \{0, 2\}$ (since $r_1 = 6 – 3 – r_3 \ge 0$).
$(3, 3, 0): \frac{6!}{3!3!0!} (1)(2^1)(1) = 20 \times 2 = 40$
$(1, 3, 2): \frac{6!}{1!3!2!} (1)(2^1)(3^1) = 60 \times 6 = 360$
$\Rightarrow$ Case 3: if $r_2 = 6$
Possible $r_3 = 0$ (since $r_1 = 6 – 6 – 0 = 0$).
$(0, 6, 0): \frac{6!}{0!6!0!} (1)(2^2)(1) = 1 \times 4 = 4$
Total Sum $= 1 + 45 + 135 + 27 + 40 + 360 + 4 = 612$.
Ans. 612