7
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Do you mean $3\cdot \sqrt{5\sqrt[3]{37} - 16} = \sqrt[3]{a} - \sqrt[3]{b} - c$? $\endgroup$ – NasuSama Feb 22 '14 at 19:14
  • 1
    $\begingroup$ Where did this problem come from? $\endgroup$ – davidlowryduda Feb 22 '14 at 19:15
  • 2
    $\begingroup$ 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 $\sqrt[3]37$ $\endgroup$ – NasuSama Feb 22 '14 at 19:16
  • 1
    $\begingroup$ 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? $\endgroup$ – Dejan Govc Feb 22 '14 at 19:32
  • 1
    $\begingroup$ Are the numbers supposed to be integers - otherwise you can take $a$ as large as you like, $b=0$ and c large too. $\endgroup$ – Mark Bennet Feb 22 '14 at 19:32
10
$\begingroup$

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 $a+b+c=1667$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Wow! Really elegant method. Thanks! $\endgroup$ – Henry Durham Feb 22 '14 at 19:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.