How to show this equals 1 without "calculations"

Question

We have $$ \sqrt[3]{2 +\sqrt{5}} + \sqrt[3]{2-\sqrt{5}} = 1 $$

Is there any way we can get this results through algebraic manipulations rather than just plugging it into a calculator?

Of course, $(2 +\sqrt{5}) + (2-\sqrt{5}) = 4 $, maybe this can in some way help?

Have you tried cubing the LHS using $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ ? After a bit of algebraic work, it comes out quite nicely. — Guillermo Angeris, Sep 18 '14 at 21:23
@GuillermoAngeris I tried, but it didn't work out, I end up with the same problem — tony, Sep 18 '14 at 21:42
@thanasissdr Well, this question arises from trying to find the roots of a cubic function. — tony, Sep 18 '14 at 22:09
See also https://math.stackexchange.com/questions/396915/how-does-one-evaluate-sqrt3x-iy-sqrt3x-iy?answertab=votes#tab-top — Watson, Dec 24 '16 at 16:27

score 5 · Answer 1 · answered Sep 18 '14 at 21:31

5

$u^3=2+\sqrt 5$, $v^3=2-\sqrt 5$

$u^3v^3=4-5=-1$ so $uv=-1$.

$u^3+v^3=4$

$(u+v)^3=u^3+v^3+3uv(u+v)=4-3(u+v)$.

That is, $u+v$ is a root of the equation $X^3+3X-4=0$ (1). The derivative of this polynomial is $3X^2+3$ which has no roots. By Rolle's theorem, that means that (1) has only one real root. You can easily check that $1$ is that root.

answered Sep 18 '14 at 21:31

ajotatxe

65,084

I don't understand. Exactly what is attained using Cardano? I haven't used Cardano. Should I (you)? – ajotatxe Sep 18 '14 at 21:44
I used Cardano to obtain a root of the equation you mentioned, and I knew beforehand that 1 was a root. What I'm trying to do now is show that the 2 roots are equivalent. – tony Sep 18 '14 at 21:45
Ah, ok. You want to know how to obtain this $1$ with no previous knowledge of that $1$ is a root. Well, Gauss called this "casus irreductibilis"... I repeat: Gauss. I don't know if there is a known method to do that. But if there is, it must be really tough. – ajotatxe Sep 18 '14 at 21:49

Kim Jong Un · Answer 2 · 2014-09-18T22:29:04.320

4

I'll expand on Guillermo's comment: let $$ a=\sqrt[3]{2+\sqrt{5}},\quad b=\sqrt[3]{2-\sqrt{5}}. $$ You want to show that $a+b=1$. Consider $$ (a+b)^3=a^3+3a^2b+3ab^3+b^3=(a^3+b^3)+3ab(a+b)=4+3\times(-1)(a+b) $$ so if you let $x=(a+b)$ , then $x$ is a real number satisfying $$ 0=x^3+3x-4=x^3-x^2+x^2-x+4x-4\\ =(x-1)(x^2+x+4)=(x-1)\left[(x+0.5)^2+3.75\right]. $$ We infer that $x=1$.

edited Sep 18 '14 at 22:29

answered Sep 18 '14 at 21:50

Kim Jong Un

14,794
1
24
49

I don't understand why $x$ is a root of that polynomial – tony Sep 18 '14 at 22:23
Because in the line before that claim, I demonstrated that $$x^3=(a+b)^3=4-3(a+b)=4-3x$$ which implies $x^3+3x-4=0$. – Kim Jong Un Sep 18 '14 at 22:24
Actually, I'll edit the solution so that the word "root" won't even be mentioned. – Kim Jong Un Sep 18 '14 at 22:28

Alexander Vigodner · Accepted Answer · 2014-09-18T22:24:35.923

3

Let $a=\sqrt[3]{2+\sqrt{5}}$ $b=\sqrt[3]{2-\sqrt{5}}$. Obviously $a b =-1$. Now let's forget what values of $a$ and $b$ and solve the system: $$ ab=-1\\ a+b=1 $$ They imply $$ a^2-a-1=0 $$ namely $a=\frac {1\pm \sqrt{5}}{2}$. So $a=\frac {1+\sqrt{5}}{2}, b=\frac {1-\sqrt{5}}{2}$. Now taking $a^3,b^3$ we can easily get $2\pm\sqrt{5}$.

edited Sep 18 '14 at 22:24

answered Sep 18 '14 at 21:55

Alexander Vigodner

2,237

How do they imply that? I can't see it – tony Sep 18 '14 at 22:04
$ab=-1\to a(1-a)=-1\to a^2-a-1=0$ – Alexander Vigodner Sep 18 '14 at 22:05
This reasoning seems circular to me, but I'm not sure. – tony Sep 18 '14 at 22:10
Knowning that $ab=-1$ I am looking for $a,b$ such that $a+b=1$. This is quadratic equation with roots easily calculated. The cube of these roots is calculated easily too. – Alexander Vigodner Sep 18 '14 at 22:11
1

I understand that $ab=-1$. But I don't understand why you can state $a+b=1$. How can you be looking for $a,b$ when you already defined them? – tony Sep 18 '14 at 22:17
Forget what I defined. I am not using the values of $a,b$ I only using the fact that $ab=-1$. And looking for solutions when $a+b=1$. It turns out out that they what we need. – Alexander Vigodner Sep 18 '14 at 22:22
No. $\sqrt[3] {(2+\sqrt{5})(2-\sqrt{5})}=\sqrt[3] {4-5}=\sqrt[3] {-1}=-1$. – Alexander Vigodner Sep 18 '14 at 22:35
Are you kidding? – Alexander Vigodner Sep 18 '14 at 22:37
yes, I am. Thanks for the help btw – tony Sep 18 '14 at 22:44

How to show this equals 1 without "calculations"

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