Find the value of $a+b+c$

Question

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?

Answer 1

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 $a+b+c=1667$.