Computing Nested Roots

It is important to note that $a$ and $b$ might have a perfect cube factor and that the expressions $\sqrt[3]{a}$ and $\sqrt[3]{b}$ need not necessarily be in simplest form. For convenience, let $a=m^3M$ and $b=n^3N$ where $m$ and $n$ are positive integers, and $M$ and $N$ are positive integers with no perfect cube factors . Then,

$\qquad \qquad \qquad \qquad 3 \cdot \sqrt{5 \sqrt[3]{37}-16}=m\sqrt[3]{M}-n\sqrt[3]{N}-c$

Squaring both sides of the equation above yields:

$\quad \qquad 45\sqrt[3]{37}-144=m^2M^{\frac{2}{3}}+n^2N^{\frac{2}{3}}+c^2-2mn\sqrt[3]{MN}-2mc\sqrt[3]{M}+2nc\sqrt[3]{N}$

Since the radicand in the LHS is $37$ , then $M$ and $N$ are expressions having one of their factor(s) as $37^x$ where $x$ is either $1$ or $2$ . If $M=k_1\cdot37^x$ and/or $N=k_2\cdot37^x$ (where $k_i$ is a positive integer), then the expression $M^{\frac{2}{3}}$ and $N^{\frac{2}{3}}$ will not have $37$ alone as the radicand unless the term $k_i^2$ is a perfect cube which is a contradiction to our assumption that $M$ and $N$ have no perfect cube factors. Therefore, $M$ and $N$ are powers of $37$ . $M$ and $N$ cannot be both equal to $37$ or both $37^2$ ; else we will have a term with radical $\sqrt[3]{37^{\frac{2}{3}}}$ which can never be found in the LHS. Therefore, one is $37$ and the other one is $37^2$ . Clearly, the only expression in the RHS that can cancel out the $\sqrt[3]{37^{\frac{2}{3}}}$ term is $-2mc\sqrt[3]{M}$ . Thus, $M=37^2$ and $N=37$ .

The left-hand side (LHS) of the last equation contains a term free of radicals ( $-144$ ). Hence, one or more quantities in the right-hand side (RHS) must also be free of radicals. At a glimpse, we note the $c^2$ is one of such terms. But then again, $c^2$ cannot be equal to $-144$ as we have assumed it initially to be a positive integer. Therefore, there is another term among the other five in the RHS that results to an integer. But which is which? Taking a closer look, we can infer that the first, second, fifth, and sixth are all radical expressions as $M$ and $N$ aren't perfect cube integers which leaves us the term $-2mn\sqrt[3]{MN}=-74mn$ as the other integer.

Summing all these conclusions, and comparing the nature of terms in LHS and RHS of the equation above and solving for the unknown variables $m,n$ and $c$ , we have:

integer term: $\quad -144=-74mn+c^2 \\ \sqrt[3]{37}: \qquad 45=37m^2+2nc \\ \sqrt[3]{37^{\frac{2}{3}}}: \qquad 0=n^2-2mc$

Substituting the expression of $c$ from the third equation to the other two and then rewriting the resulting equations:

$37m^3+n^3=45m \\ 296m^3n-n^4=576m^2$

Multiplying the first of the two by $n$ and then adding this to the second one, and finally rewriting this in favor of a constant term gives:

$m(333m-\frac{576}{n})=45$

Substituting to $m$ factors of $45$ and then finding the value of $n$ , we then get a single pair $(m,n)$ such that both $m$ and $n$ are positive integers: $m=1,n=2$ . Thus, $c=2$ .

Therefore, $a=1^3 \cdot 37^2=37^2\\ b=2^3 \cdot 37=8\cdot37$

and * $a+b+c=\boxed{1667}$ *