Radical Dude!

Letting the infinite continued fraction under the root sign be $S$ , we have that

$S = 527 - \dfrac{1}{S}$ .

(Note that we could define $S$ as the limit as $n \rightarrow \infty$ of the sequence defined recursively by $s_{0} = 527$ and $s_{n+1} = 527 - \frac{1}{s_{n}}$ . The limit exists since this sequence is decreasing, and since $s_{n} \gt 1$ for all $n$ we know that $S \ge 1$ , (both of which claims can be established by induction).)

Now let $x = \sqrt[8]{S}$ . Then $x^{8} = S$ , and so $x^{8} + \dfrac{1}{x^{8}} = 527$ .

We then observe that

$(x^{4} + \dfrac{1}{x^{4}})^{2} = x^{8} + 2 + \dfrac{1}{x^{8}} = 527 + 2 = 529$ ,

and so $x^{4} + \dfrac{1}{x^{4}} = \sqrt{529} = 23$ .

Next, we have that $(x^{2} + \dfrac{1}{x^{2}})^{2} = x^{4} + 2 + \dfrac{1}{x^{4}} = 23 + 2 = 25$ ,

and so $x^{2} + \dfrac{1}{x^{2}} = \sqrt{25} = 5$ .

Now $(x + \dfrac{1}{x})^{2} = x^{2} + 2 + \dfrac{1}{x^{2}} = 5 + 2 = 7$ ,

and so $x + \dfrac{1}{x} = \sqrt{7} \Longrightarrow x^{2} - \sqrt{7}*x + 1 = 0$

$\Longrightarrow x = \dfrac{\sqrt{7} \pm \sqrt{3}}{2}$ .

With $S \ge 1$ we know that $x \ge 1$ , so $x = \dfrac{\sqrt{7} + \sqrt{3}}{2}$ and thus

$abc = 7*3*2 = \boxed{42}$ .

sir Brian's solution is great. but we should post any different solutions. so here it is $$ first, solve $527-\dfrac{1}{527-\dfrac{1}{527-\dfrac{1}{527-\dotsm\dotsm}}} =\dfrac{527+\sqrt{525\times 529}}{2}$ now it becomes interesting. write $\sqrt[8]{\dfrac{527+\sqrt{525\times 529}}{2}}=\sqrt{\sqrt{\sqrt{\dfrac{1054+2\sqrt{525\times 529}}{4}}}}$ ] now use this $\sqrt{a+2\sqrt{b}}=\sqrt{d}+\sqrt{e}\longrightarrow a+2\sqrt{b}=d+e+2\sqrt{de}$ $a=d+e,b=de$ from what we see, lets solve this $\sqrt{\dfrac{1054+2\sqrt{525\times 529}}{4}}=\sqrt{\dfrac{525+529+2\sqrt{525\times 529}}{4}}=\dfrac{\sqrt{529}+\sqrt{525}}{2}=\dfrac{23+\sqrt{525}}{2}$ repeat the step $\sqrt{\dfrac{23+\sqrt{525}}{2}}=\sqrt{\dfrac{25+21+2\sqrt{25\times 21}}{4}}=\dfrac{\sqrt{25}+\sqrt{21}}{2}=\dfrac{5+\sqrt{21}}{2}$ repeat the step for one final time $\sqrt{\dfrac{5+\sqrt{21}}{2}}=\sqrt{\dfrac{3+7+2\sqrt{3\times 7}}{4}}=\boxed{\dfrac{\sqrt{3}+\sqrt{7}}{2}}$