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 2019-09-30, 13:05 #1 wildrabbitt   Jul 2014 2·13·17 Posts 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.
2019-09-30, 14:32   #2
Dr Sardonicus

Feb 2017
Nowhere

2×23×71 Posts

Quote:
 Originally Posted by wildrabbitt 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

 2019-09-30, 14:40 #3 wildrabbitt   Jul 2014 2·13·17 Posts 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
 2019-09-30, 15:06 #4 Chris Card     Aug 2004 100000012 Posts 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
2019-09-30, 15:20   #5
Dr Sardonicus

Feb 2017
Nowhere

2×23×71 Posts

Quote:
 Originally Posted by wildrabbitt 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 $(3 + \frac{11}{3}\sqrt{3})^{3}$

2019-09-30, 19:19   #6
wildrabbitt

Jul 2014

2·13·17 Posts

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

2019-09-30, 23:07   #7
Dr Sardonicus

Feb 2017
Nowhere

2·23·71 Posts

Quote:
 Originally Posted by wildrabbitt 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

 2019-10-01, 13:55 #8 wildrabbitt   Jul 2014 1BA16 Posts 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.
2019-10-02, 04:30   #9
LaurV
Romulan Interpreter

Jun 2011
Thailand

32×953 Posts

Quote:
 Originally Posted by Dr Sardonicus 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

 2019-10-02, 08:44 #10 wildrabbitt   Jul 2014 44210 Posts 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.
2019-10-02, 14:34   #11
Dr Sardonicus

Feb 2017
Nowhere

CC216 Posts

Quote:
 Originally Posted by wildrabbitt 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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