# Writing $3.8473221018630726$ in the form $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$.

I attempted the following question on Brilliant which has to do with finding roots of a cubic polynomial. I was successful in finding what the only real root is but I am facing a problem rewriting the root in the sought expression.

The equation $$x^3-3x^2-3x-1=0$$ has exactly one real solution that can be written in the form $$\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$$. What is the value of $$a+b+c$$?

I've found the value of $$x$$ to be equal to $$3.8473221018630726$$ by the method of depressing the cubic. Any hints to proceed are appreciated.

Edit:

Would the fact that $$x=\dfrac{2}{\sqrt[3]{4}}+\sqrt[3]{4}+1=\dfrac{2}{\sqrt[3]{2}}+\sqrt[3]{2}+1\approx3.8473221018630726$$ help somewhere in determining $$a, b$$ and $$c$$?

• I, er.. think you should not resort to numerical methods.. Commented Apr 29, 2019 at 7:07
• Anyway, $a,b,c$ might not be integers, so it's hard to tell except by solving algebraically. Commented Apr 29, 2019 at 7:08
• $\frac{2}{\sqrt[3]{4}} = \frac{\sqrt[3]{8}}{\sqrt[3]{4}} = \sqrt[3]{2}$ Commented Apr 29, 2019 at 7:14
• Thanks @achillehui :) Commented Apr 29, 2019 at 7:15
• Now you know what $a,b,c$ is. A cheaper way to derive the root is $$x^3 - 3x^2 - 3x - 1 = 0 \iff 2x^3 - (x+1)^3 = 0 \iff \left(\frac{x+1}{x}\right)^3 = 2$$ For the real root, you can take cubic root on both sides and get $$\frac{x+1}{x} = \sqrt[3]{2} \implies x = \frac{1}{\sqrt[3]{2} - 1} = \frac{(\sqrt[3]{2})^2 + \sqrt[3]{2} + 1}{(\sqrt[3]{2})^3-1} = \sqrt[3]{4} + \sqrt[3]{2} + \sqrt[3]{1}$$ Commented Apr 29, 2019 at 7:27

Would the fact that $$x=\frac{2}{\sqrt[3]{4}}+\sqrt[3]{4}+1=\frac{2}{\sqrt[3]{2}}+\sqrt[3]{2}+1\approx3.8473221018630726$$ help somewhere in determining $$a, b$$ and $$c$$?

Yes, it would very much. If $$\frac{2}{\sqrt[3]{2}}+\sqrt[3]{2}+1$$ is the root you're after, then $$\frac{2}{\sqrt[3]{2}}+\sqrt[3]{2}+1 = \sqrt[3]{\frac82} + \sqrt[3]{2} + \sqrt[3]{1}$$ and you have the solution.

• Thanks for answering :) Commented Apr 29, 2019 at 7:21

From the numerical solution (which is pretty "small"), you can try to check a few small values for $$a,b,c$$... which I did:

$$3.8473221018630726=\sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4}$$

EDIT: Actually this had to be solved in the following way (for one real solution):

$$x^3 - 3x^2 - 3x - 1 = 0$$

$$2x^3 - (x+1)^3 = 0$$

$$2x^3 = (x+1)^3$$

$$\sqrt[3]2x = x+1$$

$$x = \frac1{\sqrt[3]2-1}$$

$$x = \frac1{\sqrt[3]2-1}\frac{\sqrt[3]{2^2}+\sqrt[3]{2}+1}{\sqrt[3]{2^2}+\sqrt[3]{2}+1}=\sqrt[3]{4}+\sqrt[3]{2}+\sqrt[3]1$$

• @achillehui Thanks, I have corrected the typo.
– Saša
Commented Apr 29, 2019 at 7:12
• Thanks for answering :) Commented Apr 29, 2019 at 7:15