TITLE: Find the value of $a+b+c$
QUESTION [7 upvotes]: If $$3.\sqrt{5.\sqrt[3]{37}-16}=\sqrt[3]a-\sqrt[3]b-c$$
What is the value of $a+b+c$?
EDIT: I forgot to mention that $a$, $b$ and $c$ are positive integers.

I tried squaring and then cubing but it got very lengthy. Is there some elegant method to do it?

REPLY [10 votes]: Let $\theta=\sqrt[3]{37}$. If we put $\alpha=\theta^2-2\theta-2 \approx 2.4$, then
$$
\begin{array}{lcl}
\alpha^2 &=& (\theta^2-2\theta-2)^2 \\
&=&  \theta^4 - 4 \theta^3 + 8 \theta + 4 \\
&=&  37\theta -4\times 37+8\theta+4 \\
&=& 45\theta -144 \\
&=& 9(5\theta-16)
\end{array}
$$
It follows that $3\sqrt{5\theta-16}=\alpha$, so
$a=37^2,b=8\times 37,c=2$, and hence $a+b+c=1667$.