is it necessary to densest?

$\sqrt{98}= 7\sqrt{2}$ there is 'only' one way to add 7 such that $a>b>c>0$ $4+2+1=7$ so $7\sqrt{2}=4\sqrt{2}+2\sqrt{2}+\sqrt{2}$ $\sqrt{98}=\sqrt{32}+\sqrt{8}+\sqrt{2}$ $a+b+c = 32+8+2 = \boxed{42}$

sqrt(98) = sqrt(a) + sqrt(b) + sqrt(c), a>b>c>0 and abc are integers. Find sum of a, b and c.

Solution : sqrt(98) = 7sqrt(2) = 4sqrt(2) + 2sqrt(2) + 1sqrt(2) =sqrt(2 * 4 4) + sqrt(2 * 2 2) + sqrt(2 * 1*1) = sqrt(32) + sqrt(8) + sqrt(2)

Let a = 32, b = 8, c = 2 So a>b>c>0 and abc = 32 * 8 * 2 = 512, which is an integer.

a + b + c = 32 + 8 + 2 = 42 Hence, the required sum is 42.