Math is Radical 4

First, express $18$ as $b - a$ . Then square $-a + 12\sqrt{2} + 6\sqrt{5} + 4\sqrt{10}$ , and afterwards you will obtain $a^2 + 628 + (240 - 24a)\sqrt{2} + (192 - 12a )\sqrt{5} + (144 - 8a)\sqrt{10}$ . To remove the $\sqrt{2}$ , set its coefficient in the expansion, which is $240 - 24a$ , to $0$ . That is, you will solve the linear equation $240 - 24a = 0$ for $a$ and get $a = 10$ . Going back to $b - a = 18$ , it is then clear that $b = 28$

I tried to obtain other solutions by eliminating first the $\sqrt{5}$ or the $\sqrt{10}$ ; the results are, respectively, $\sqrt{34+\sqrt{884-\sqrt{44032-\sqrt{424673280}}}}$ and $\sqrt{36+\sqrt{952-\sqrt{76608+\sqrt{849346560}}}}$ . The problem with these results is that some of the radicals are preceeded by a minus sign, whereas the question specifies a representation containing only plus signs.

The purpose of introducing $28 - 10$ is to eliminate the $\sqrt{2}$ in the expansion. If we merely square $12\sqrt{2} + 6\sqrt{5} + 4\sqrt{10}$ , the expansion will still show a term with a $\sqrt{2}$ , since

$(12\sqrt{2} + 6\sqrt{5} + 4\sqrt{10})^{2}$ = $628 + 240\sqrt{2} + 192\sqrt{5} + 144\sqrt{10}$ ,

which will not help us, since in the succeeding ''squarings'', we want to eliminate the radicals $\sqrt{2} , \sqrt{5} , and \sqrt{10}$ one by one.