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Old 2019-09-30, 13:05   #1
wildrabbitt
 
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Default root given by del Ferro's formula

Can anyone help me prove this equals 6?


\(\sqrt[3]{390+\sqrt{\frac{4937284}{3\sqrt{3}}}}\)-\(\sqrt[3]{-390+\sqrt{\frac{4937284}{3\sqrt{3}}}}\)


I'd like to be able to without knowing already work out what an expression like this equals if it equals an integer.
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Old 2019-09-30, 14:32   #2
Dr Sardonicus
 
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Quote:
Originally Posted by wildrabbitt View Post
Can anyone help me prove this equals 6?


\(\sqrt[3]{390+\sqrt{\frac{4937284}{3\sqrt{3}}}}\)-\(\sqrt[3]{-390+\sqrt{\frac{4937284}{3\sqrt{3}}}}\)


I'd like to be able to without knowing already work out what an expression like this equals if it equals an integer.
What you posted looks like


\sqrt[3](390 + \sqrt{\frac{4937284}{3\sqrt{3}}}}) - \sqrt[3](-390 + \sqrt{\frac{4937284}{3\sqrt{3}})

which doesn't fit the formula for a cubic in Q[x] and isn't 6.

Please post the cubic polynomial. Thank you.

Last fiddled with by Dr Sardonicus on 2019-09-30 at 14:38 Reason: xinfig posty
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Old 2019-09-30, 14:40   #3
wildrabbitt
 
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Default first post revised

Thanks. Before I accept the fact that I can't write a post without messing it up in some way and give up, I'll try and write out what I should have written.


I'd like to be able to understand how to prove something like



\(\sqrt[3]{390+\frac{1}{3}\sqrt{\frac{4937284}{3}}}-\sqrt[3]{-390+\frac{1}{3}\sqrt{\frac{4937284}{3}}}\)=6




Sorry. The polynomial. \(y = x^3+94x-780\)

Last fiddled with by wildrabbitt on 2019-09-30 at 14:47 Reason: runaway braces
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Old 2019-09-30, 15:06   #4
Chris Card
 
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I asked a similar question on math.stackexchange a while ago: https://math.stackexchange.com/quest...ining-radicals

Some of the replies there might help

Chris
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Old 2019-09-30, 15:20   #5
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Quote:
Originally Posted by wildrabbitt View Post
Thanks. Before I accept the fact that I can't write a post without messing it up in some way and give up, I'll try and write out what I should have written.


I'd like to be able to understand how to prove something like



\(\sqrt[3]{390+\frac{1}{3}\sqrt{\frac{4937284}{3}}}-\sqrt[3]{-390+\frac{1}{3}\sqrt{\frac{4937284}{3}}}\)=6




Sorry. The polynomial. \(y = x^3+94x-780\)
Well, assuming you have correctly applied the formulas to this polynomial (and your retyped expression does evaluate, to a fare-thee-well, to 6), the proof is simple: Your expression is real, and the cubic is the product of x - 6 and x^2 + 6*x + 130, a quadratic with negative discriminant.

But starting with your expression, and from there proving it is equal to 6, the key is that expression

390 + \sqrt{\frac{4937284}{27}} \; = \;

390 + \frac{2\cdot 11\cdot 101}{9}\sqrt{3}

is equal to (3 + \frac{11}{3}\sqrt{3})^{3}
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Old 2019-09-30, 19:19   #6
wildrabbitt
 
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Thanks for factoring 4937284 and dividing the polynomial by (x-6) and helpfully explaining what I can do.



Quote:
Well, assuming you have correctly applied the formulas to this polynomial (and your retyped expression does evaluate, to a fare-thee-well, to 6), the proof is simple: Your expression is real, and the cubic is the product of x - 6 and x^2 + 6*x + 130, a quadratic with negative discriminant.
So this way, prove (x-6) divides the polynomial with remainder 0. Prove the remainder has two complex roots so there must be one real one so the real one must be 6. Then since Cardano's formula yields a real root the root it yields must be 6. QED


The one with the expression was what I was more interested in understanding. I suppose I'll never know how you knew to do that.
I guess it's the case that if Cardano's formula results in an integer value like.
that, the expression inside the cube roots are always cubes.


Thanks very much, and to Chris.

Last fiddled with by wildrabbitt on 2019-09-30 at 19:19
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Old 2019-09-30, 23:07   #7
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Quote:
Originally Posted by wildrabbitt View Post
The one with the expression was what I was more interested in understanding. I suppose I'll never know how you knew to do that.
I guess it's the case that if Cardano's formula results in an integer value like.
that, the expression inside the cube roots are always cubes.
Oh, but I'm happy to explain. In the first instance, of course, factoring the polynomial is the first thing to try.

But you're right, the only way the formula can give an integer is if the expression inside the cube roots is a perfect cube. And, if you assume the expressions under the cube roots are the cubes of expressions of the form a \; + \; \sqrt{b}, then in your case "a" has to be 3. This is a great help in extracting the cube root.

But the factorization of the cubic is what drove me to look closer at that square root. You see, the formulas will also give the roots of the quadratic factor. And I know that (for a "reduced" cubic whose x2 term is 0, like yours) the formulas look like

\sqrt[3]{a + \sqrt{b}} \; + \; \sqrt[3]{a - \sqrt{b}}\text{ , }\omega\cdot\sqrt[3]{a + \sqrt{b}} \; + \; \omega^{2}\cdot\sqrt[3]{a - \sqrt{b}}\text{, and }\omega^{2}\cdot\sqrt[3]{a + \sqrt{b}} \; + \; \omega\cdot\sqrt[3]{a - \sqrt{b}}\text{, where }\omega^{2} \; + \; \omega \; + \; 1 \; = \; 0\text{.}

Now the discriminant of the quadratic is -484, whose square root is 22i. So multiplying the primitive cube roots of 1 by an expression of the form a \; + \; \sqrt{b} has to yield a term that is a rational multiple of i. Now the cube roots of 1 involve the square roots of -3, so the expressions under the cube roots have to involve the square roots of 3. Factoring confirmed this.

And once I had the expression 390 + 2222/9 * sqrt(3), and knew the cube root had to be of the form 3 + c*sqrt(3) in order to get the two cube roots to add up to exactly 6, it was easy to find the coefficient c.

You can also use software to either find a "quadratic surd cube root of a quadratic surd" or show there isn't one. In your case, 390 + 2222/9*sqrt(3),

Code:
? lift(factornf(x^3 - 390 - 2222*t/9, t^2 - 3))

%1 = 
[x + (-11/3*t - 3) 1]

[x^2 + (11/3*t + 3)*x + (22*t + 148/3) 1]
and -- ta-daaah! -- x = 3 + 11/3* sqrt(3). If there's no cube root of the right form, the cubic remains irreducible.

Square roots of quadratic surds are even more fun, because they can "come out exact" with expressions that involve more than one square root. For example,

\(\frac{\sqrt{2}+\sqrt{6}}{2}\)^2 \; =\; 2 \; + \; \sqrt{3}

I know there's a classic (read "old") paper on simplifying square roots of quadratic surds, but at the moment I can't recall where I read the reference to it...

Last fiddled with by Dr Sardonicus on 2019-09-30 at 23:15 Reason: fignix spoty
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Old 2019-10-01, 13:55   #8
wildrabbitt
 
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Thanks very much. I'm so glad I know now.



Hencewith I'm wondering three things.


One is what I would do if I hadn't known the real root was 6 as you seem to have needed that fact to work out the value of c.


Another is in which thread it was that I asked you what that language is you use.


The third I think I shouldn't mention unless I figure out something myself.
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Old 2019-10-02, 04:30   #9
LaurV
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Quote:
Originally Posted by Dr Sardonicus View Post
Square roots of quadratic surds are even more fun, because they can "come out exact" with expressions that involve more than one square root. For example,

\(\frac{\sqrt{2}+\sqrt{6}}{2}\)^2 \; =\; 2 \; + \; \sqrt{3}
They always do, you amplify with the conjugate. Here in this particular case you don't need to amplify, because your expression can simplify by \(\sqrt 2\).


\[\left(\frac{\sqrt 2+\sqrt 6}2\right)^2 \; =\; \left(\frac{\sqrt 2+\sqrt {2\cdot 3}}2\right)^2 \; =\; \left(\frac{\sqrt 2+\sqrt 2\cdot\sqrt 3}2\right)^2 \; =\; \left(\frac{\sqrt 2\cdot(1+\sqrt 3)}2\right)^2 \; =\; \left(\frac{1+\sqrt 3}{\sqrt 2}\right)^2 \; =\; \frac{(1+\sqrt 3)^2}{(\sqrt 2)^2} \; =\; \frac{1+2\cdot\sqrt 3+3}{2}\; =\; \frac{4+2\cdot\sqrt 3}{2}\;=\;2+\sqrt 3\]


Edit: of course, that is a very roundabout way to "solve" it, like scratching your left temple with the right hand around your nape. But I tried to explain the "why".

Last fiddled with by LaurV on 2019-10-02 at 04:37
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Old 2019-10-02, 08:44   #10
wildrabbitt
 
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That particular one only works because \(\sqrt{6}\) has \(\sqrt{2}\) as a factor so the numerator can be factorised.
I don't see how it would work if \(\sqrt{19}\) had been the root.
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Old 2019-10-02, 14:34   #11
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Quote:
Originally Posted by wildrabbitt View Post
That particular one only works because \(\sqrt{6}\) has \(\sqrt{2}\) as a factor so the numerator can be factorised.
I don't see how it would work if \(\sqrt{19}\) had been the root.
Such examples may be found by substituting x^2 for x in a quadratic polynomial (assumed to be irreducible), e.g. x^2 - 340*x + 1. This produces a quartic, x^4 - 340*x^2 + 1. If the quartic is irreducible with Galois group E(2), the "four-group" [as it is in this example], the roots of the quadratic in x will have a factorization with two square roots:

170 + 39\sqrt{19} \; = \; \(\frac{13\sqrt{2} \; + \; 3\sqrt{38}}{2}\)^{2}

BTW I use Pari-GP.
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