Mathematics

As I said in a previous answer, finding out that such a simplification occurs is exactly as hard as finding out that $35+18\sqrt{-3}$ has a cube root in $\Bbb Q(\sqrt{-3})$ (well actually, $\Bbb Z[(1+\sqrt{-3})/2]$ because $35+18\sqrt{-3}$ is an algebraic integer). Suppose $p,q,d$ are integers. ...

Twink

lol

xD

it was more expensive

user537566

7:07 PM

xD

Twink

@mercio Suppose I'm a high school student...

I don't know what $i$ means...

mercio

$i$ is $\sqrt{-1}$ but it doesn't really matter for the methods

Twink

and to prove that $\sqrt[3]{9+4 \sqrt{5}}+\sqrt[3]{9-4 \sqrt{5}}=3$?

we go back to the first problem

mercio

you guess that the cube root is of the form $3/2 + b \sqrt 5$, so you cube that and solve for $b$ and you find a rational solution for $b$ and all is well

Twink

If I do that I can get $a$ and $b$ and get $\sqrt[3]{9+4 \sqrt{5}}$ directly

but I don't wanna suppose that, why should I suppose that?

mercio

7:18 PM

you should because it is a result of algebraic number theory that if the sum is equal to $3$ (or any rational number) then there is a cube root of the form $3/2 + b\sqrt 5$ with $b$ rational

I don't think there is any method for this kind of thing that will not seem cheat-y to a high schooler

Twink

but I don't know if the sum is 3 nor if $\sqrt[3]{9+4 \sqrt{5}}$ is of the form $a+b\sqrt{5}$

user537566

Try to multiply by conjugate.

Twink

O.O

mercio

how exactly is your problem stated ?

Twink

expand $\sqrt[3]{9+4 \sqrt{5}}$?

mercio

7:26 PM

o. .O

then you use a meta-theorem that says that a hig hschool exercise always have nice solutions

so you look at the sum and conjecture that it is $3$ then guess the cube root is of the form blablabla

and actually, it is not that easy to find $a$ and $b$ such that $(a+b\sqrt 5)^3 = 9+4\sqrt 5$. However it becomes easy if you know that $a=3/2$

Daminark

@Eric finally done with GMT!

Twink

ok I'm gonna do it that way thanks

and how can I solve the equation $x^3-3x-18=0$?

mercio

you use a meta-theorem that says that high school exercises have nice solutions

and you find $x=3$

(by trying every integer until you find a solution)

Twink

lol ¬¬

but with a method

mercio

actually you should use the rational root theorem

it limits the possible cases

(and you still have to assume that there is a rational root)

user537566

7:40 PM

Let $x= \sqrt[3]{9+4 \sqrt{5}}$ and $y=\sqrt[3]{9-4 \sqrt{5}}$. Find $(x+y)^3$ and $(x-y)^3$.

mercio

basically if there is no rational root you would be totally screwed

and they wouldn't ask that to a high schooler

Twink

now suppose I'm not in highschool

mercio

then you are still totally screwed if there is no rational root

and i suppose you could use Cartan's formulas but you will still hate them

user537566

You will find that $(x+y)^3 = 27$ i.e. $x+y = 3$.

Twink

but there is one

mercio

7:43 PM

the thing is that cartan's formula is actually useless

because it doesn't show you when there is a simple solution

@user537566 that supposes you are able to solve $x^3-3x-18 = 0$

how do i solve $x^3 - 3x -18 = 0$ ? with cartan's formula ! wait it tells me to simplify an ugly sum of cube root ! how do i simplify the cube roots ? by solving $x^3-3x-18 = 0$ ! yayyyyy !

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$Mathematics$ Mathematics