Find a sum of the rational terms in the following expression after full expanding. $$\left(\sqrt{2}+\sqrt{27}+\sqrt{180}\right)^{10}$$
Since the term should be rational, each power should be even.
Number of non negative even integers $m,n,q$ satisfying
$$m+n+q=10$$ gives number of triplets as $21$..but its very tedious to find sum of $21$ terms...any other approach?
On
Using Galois theory (as in pisco125's asnwer) and a pocket calculator, $$\tag1\begin{align} (\sqrt 2+3\sqrt 3+12\sqrt5)^{10}&\approx 10377910865876.76\\ (\sqrt 2+3\sqrt 3-12\sqrt5)^{10}&\approx 213278054.90\\ (\sqrt 2-3\sqrt 3+12\sqrt5)^{10}&\approx 6890930010.35\\ (\sqrt 2-3\sqrt 3-12\sqrt5)^{10}&\approx 2263950893293.99\\ \end{align}$$ The average of these four numbers is very close to the integer $$3162241491809.$$ Strictly speaking, we should take the average of eight numbers, namely also those with a negative sign for $\sqrt 2$, but due to the even exponent, we need only half of the summands. As we know from the theory that the answer must be an integer, only decent precision is required in the computations $(1)$.
On
Let $a=\sqrt2$, $b=3\sqrt3$ and $c=6\sqrt5$.
Hence, the needed number is $$\sum_{cyc}\left(a^{10}+\binom{10}{8}(a^8b^2+a^8c^2)+\binom{10}{6}(a^6b^4+a^6c^4)+\binom{10}{6,2,2}a^6b^2c^2+\binom{10}{4,4,2}a^4b^4c^2\right)=$$ $$=188971148939+45\cdot30539765574+210\cdot4912928964+$$ $$+1260\cdot322052760+3150\cdot51263280=3162241491809.$$
Note that the Galois group of $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ over $\mathbb{Q}$ is generated by three automorphism: one swaps $\sqrt{2}$ to $-\sqrt{2}$, the other swaps $\sqrt{3}$ and $-\sqrt{3}$, the last one swaps $\sqrt{5}$ and $-\sqrt{5}$.
Let $\alpha_1,\alpha_2,\alpha_3,\alpha_4$ be the orbits of $(\sqrt{2}+3\sqrt{3}+6\sqrt{5})^2$ under the action of Galois group.
Note that ${\alpha_1}^{5}+{\alpha_2}^{5}+{\alpha_3}^{5}+{\alpha_4}^{5}$ is a rational number, because it is fixed by the three automorphisms above. Your desired answer is $$ \frac{{\alpha_1}^{5}+{\alpha_2}^{5}+{\alpha_3}^{5}+{\alpha_4}^{5}}{4}$$ If your acumen is strong enough, you can also arrive at this result without any Galois theory. Also note that $\alpha_i$ are roots of the root of the polynomial $$510082225 - 19503140 x + 219894 x^2 - 836 x^3 + x^4 = 0 $$ Repeated application of Netwton's identities gives the value of your sum.
As an alternative, a recurrence relation can be used to compute $$S_n = {\alpha_1}^n + {\alpha_2}^n + {\alpha_2}^n + {\alpha_3}^n + {\alpha_4}^n$$ which is given by $$510082225 S_n- 19503140 S_{n+1} + 219894 S_{n+2} - 836 S_{n+3} + S_{n+4} = 0 $$ valid for $n\geq 1$. You have to know $S_1,S_2,S_3$ to get this recurrence running. You might want to use this as an alternative to compute $S_5$ or $S_n$ for higher $n$.