Find the value of $a+b+c$

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?

• Do you mean $3\cdot \sqrt{5\sqrt[3]{37} - 16} = \sqrt[3]{a} - \sqrt[3]{b} - c$? – NasuSama Feb 22 '14 at 19:14
• Where did this problem come from? – davidlowryduda Feb 22 '14 at 19:15
• When typing more than a character under the square root bracket, be sure to type for instance \sqrt[3]{37} instead of \sqrt[3]37. \sqrt[3]{37} gives $\sqrt[3]{37}$ whereas \sqrt[3]37 gives $\sqrt[3]37$ – NasuSama Feb 22 '14 at 19:16
• I don't think this problem is well-posed. For example, you might take $a=0,b=0,c=-3\sqrt{5\sqrt[3]{37} - 16}$ or $a=(3\sqrt{5\sqrt[3]{37} - 16})^3,b=0,c=0$, which both satisfy the equation (unless I'm missing something), but the sums $a+b+c$ differ: one is negative and one is positive. Should we assume that $a,b,c$ are integers or something like that? – Dejan Govc Feb 22 '14 at 19:32
• Are the numbers supposed to be integers - otherwise you can take $a$ as large as you like, $b=0$ and c large too. – Mark Bennet Feb 22 '14 at 19:32

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 $a+b+c=1667$.