The Roots of the Problem

The above radical equation is equivalent to $\large x^{1/a + 1/b + 1/c} = x^{1/3}$ , or:

$\large \frac{1}{c} = \frac{1}{3} - \frac{1}{a} - \frac{1}{b} \Rightarrow c = \frac{3ab}{ab - 3(a+b)}$ (i).

The maximum value of $c$ will occur when the denominator in (i) is a minimum. If $a,b,c \in \mathbb{N}$ , then the minimum denominator shall equal $1$ , or;

$\large ab - 3a - 3b = 1 \Rightarrow b = \frac{3a+1}{a-3} = 3 + \frac{10}{a-3}$ (ii)

which has the solution set in positive integers $(a,b) = (4,13); (5,8); (8,5); (13,4)$ . Since $a \le b$ , we only admit $(a,b) = (4,13); (5,8).$ If we substitute these two pairs into (i), then we obtain:

$\large c = \frac{3(4)(13)}{1} = 156$ or $\large c = \frac{3(5)(8)}{1} = 120$

Hence, $\boxed{c_{MAX} = 156}.$

Same idea as @Carsten Meyer 's solution here .

1/3 = 1/(3+1) + 1/(3(3+1))
= 1/4 + 1/12
1/12 = 1/(12+1) + 1/(12(12+1))
= 1/13 + 1/156

1/3 = 1/4 + 1/13 + 1/156