20190930, 13:05  #1 
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. 
20190930, 14:32  #2  
Feb 2017
Nowhere
2^{2}·821 Posts 
Quote:
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 20190930 at 14:38 Reason: xinfig posty 

20190930, 14:40  #3 
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. \(y = x^3+94x780\) Last fiddled with by wildrabbitt on 20190930 at 14:47 Reason: runaway braces 
20190930, 15:06  #4 
Aug 2004
3×43 Posts 
I asked a similar question on math.stackexchange a while ago: https://math.stackexchange.com/quest...iningradicals
Some of the replies there might help Chris 
20190930, 15:20  #5  
Feb 2017
Nowhere
110011010100_{2} Posts 
Quote:
But starting with your expression, and from there proving it is equal to 6, the key is that expression is equal to 

20190930, 19:19  #6  
Jul 2014
1BA_{16} Posts 
Thanks for factoring 4937284 and dividing the polynomial by (x6) and helpfully explaining what I can do.
Quote:
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 20190930 at 19:19 

20190930, 23:07  #7  
Feb 2017
Nowhere
2^{2}·821 Posts 
Quote:
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 , 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 x^{2} term is 0, like yours) the formulas look like 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 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] 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, 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 20190930 at 23:15 Reason: fignix spoty 

20191001, 13:55  #8 
Jul 2014
110111010_{2} 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. 
20191002, 04:30  #9  
Romulan Interpreter
Jun 2011
Thailand
2^{2}×7×307 Posts 
Quote:
\[\left(\frac{\sqrt 2+\sqrt 6}2\right)^2 \; =\; \left(\frac{\sqrt 2+\sqrt {2\cdot 3}}2\right)^2 \; =\; \left(\frac{\sqrt 2+\sqrt 2\cdot\sqrt 3}2\right)^2 \; =\; \left(\frac{\sqrt 2\cdot(1+\sqrt 3)}2\right)^2 \; =\; \left(\frac{1+\sqrt 3}{\sqrt 2}\right)^2 \; =\; \frac{(1+\sqrt 3)^2}{(\sqrt 2)^2} \; =\; \frac{1+2\cdot\sqrt 3+3}{2}\; =\; \frac{4+2\cdot\sqrt 3}{2}\;=\;2+\sqrt 3\] 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 20191002 at 04:37 

20191002, 08:44  #10 
Jul 2014
2×13×17 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. 
20191002, 14:34  #11  
Feb 2017
Nowhere
CD4_{16} Posts 
Quote:
BTW I use PariGP. 

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