simplify $\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$

Question

simplify $$\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$$

1.$\frac{3}{2}$

2.$\frac{\sqrt[3]{65}}{4}$

3.$\sqrt[3]{2}$

4.$1$

I equal it to $\sqrt[3]{a}+\sqrt[3]{b}$ but I cant find $a$ and $b$

No it isn't a complex number the first one is a real one als the second one let me make it bigger.The first one is around $\sqrt{7}$ and the second one is around $-1$ — Taha Akbari, Sep 01 '16 at 09:24
http://www.wolframalpha.com/input/?i=(5%2B2+sqrt(13))%5E(1%2F3)-%7B-5%2B2%5Csqrt%7B13%7D%7D%5E(1%2F3) You get 1. — Watson, Sep 01 '16 at 09:31
@JanEerland: the cubic root of a is the equal to $$-\sqrt[3]{-a}$$ — Watson, Sep 01 '16 at 09:33
What it all this talk about complex numbers? It seems completely inappropriate here, it is a cubic root, not a square one — Vincent, Sep 01 '16 at 09:34
You have been asking many questions today. Don't you think you can try to work by your own as well ? — Jean Marie, Sep 01 '16 at 09:43

Piquito · Accepted Answer · 2016-09-01T09:53:36.363

8

Let $X=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$ From $(A+B)^3=A^3+B^3+3AB(A+B)$ we get the equation $$X^3-3ABX-(A^3+B^3)=0$$ We have $(AB)^3=({5+2\sqrt{13}})({5-2\sqrt{13}})=-27\Rightarrow {AB}=-3$ and $A^3+B^3=10$. Hence our equation $$X^3+9X-10=0\iff (X-1)(X^2+X+10)=0$$ Thus $$X=1$$ It is the $4$ the asked simplification.

edited Sep 01 '16 at 09:53

answered Sep 01 '16 at 09:33

Piquito

29,594

Just a minor typo: you meant $\sqrt[3]{-27}$. – String Sep 01 '16 at 09:42
Thank you very much. It was in fact a typo I edit. – Piquito Sep 01 '16 at 09:46

score 2 · Answer 2 · answered Sep 01 '16 at 09:32

2

Note your number $\alpha$

It turns out that (after some calculation) that $\alpha^3=10-9\alpha$

1 seems to do the trick...

answered Sep 01 '16 at 09:32

Vincent

2,318

There is no reason for the downvote. Note that this answer is basically the same as mine. I put an upvote. – Piquito Sep 01 '16 at 09:41

score 0 · Answer 3 · answered Sep 01 '16 at 09:39

Hint: Rewrite $x_{1,2}=5\pm 2\sqrt{13}$ in polar coorinates $x=|x|\exp(i\phi_{1,2})$. Determine $|x_{1,2}|=\sqrt{5^2+(2\sqrt{13})^2}$ and $\tan(\phi_{1,2})=\frac{\pm 2\sqrt{13}}{4}$

Use the periodicity of the exponential function in the complex domain $\sqrt[3]{x_{1,2}}=|x_{1,2}|^{1/3}\exp(i\phi_{1,2}/3+2\pi k/3)$, in which $k \in \mathbb{Z}$ (you will have to check which range of values for $k$ give new solutions.

simplify $\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$

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