An Olympiad Question on cube roots.

Question

If $$\sqrt[3]{\sqrt[3]{2}-1} = \sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{\frac{1}{9}}$$where $a$ is greater than $b$, then find the value of $a+2b+3$.

I got this question on an olympiad handout and couldn't figure out , how to attempt this . I tried cubing , but to no avail . Please help!

Did your handout give the particular olympiad, year and problem number of the question? — coffeemath, Apr 07 '23 at 14:55
It is also important if there is a restriction on $a,b$ such as their being integers, or perhaps rationals. [There are many real solutions] — coffeemath, Apr 07 '23 at 15:08
You say you tried cubing, what happens when you do that and why do you think it's fruitless? Please show some of your work. — user170231, Apr 07 '23 at 17:19
I've seen this problem before. It's from ARML, the last problem of an individual round in the 1990s. I don't think anyone got it right. — TheBestMagician, Apr 07 '23 at 18:19
Related: Denesting radicals like $\sqrt[3]{\sqrt[3]{2} - 1}$. — dxiv, Apr 07 '23 at 19:02

TheBestMagician · Accepted Answer · 2023-04-07T18:29:14.220

I managed to recover the old ARML problem.

1997 ARML I-8: If $\sqrt[3]{\sqrt[3]{2}-1}$ is written as $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c},$ where $a,b,$ and $c$ are rational numbers, compute the sum $a+b+c.$

It suffices to solve this problem. First I claim that the following is true: $$\sqrt[3]{\sqrt[3]{2}-1}=\frac{\sqrt[3]{3}}{\sqrt[3]{2}+1}$$ This can be found by setting $k=\sqrt[3]{2}$ and manipulating $k^3-2=0$ to somehow get a $k-1$ term and a $(k+1)^3$ term. With some creativity, we can find $(k-1)(k+1)^3=3$ (this is not hard to verify), which shows the above claim.

Finishing the problem from here is easy. Simply multiply the denominator and numerator by $\sqrt[3]{4}-\sqrt[3]{2}+1$ (motivated by the factorization of $k^3+1$).

Note: This identity was discovered by Ramanujan a very long time ago. See proposition 2.6 of this paper.

An Olympiad Question on cube roots.

1 Answers1