On 2020-07-11 15:56, Henry Baker wrote:
Here's one method that a high schooler could follow, but perhaps might not be able to develop himself/herself.
Reverse engineer the given expression to guess that it is Cardano's formula for the real root of the cubic equation x^3+6*x-20.
Then factor this equation itself into x^3+6*x-20 = (x-2)*(x^2+2*x+10) to prove that 2 is the real root, hence the original given expression equals 2.
I'm not at all happy with this explanation, as it pulls a deus ex machina rabbit out of the hat.
Can someone provide a better-motivated solution?
If you just cube the original equation (which seems natural given the cube root), and cancel terms, you get 20 - 6(sqrt(108) + 10)^(1/3) + 6(sqrt(108) - 10)^(1/3) If x denotes (sqrt(108)+10)^(1/3)-(sqrt(108)-10)^(1/3), the original term, then x^3 = 20 - 6x, so the x^3 + 6x - 20 = 0 seems to fall out "naturally" to me. Maybe I am missing what you think is being pulled out of a hat here. I am trying to channel my early-teen self, and I think cubing the expression and equating it to x^3 would have been one of the first things I would have tried. (I would also have calculated the first few digits of the numerical value of the original expression, so I would have had a guess as to what the value was, so I am pretty sure I would have guessed that x-2 was a factor of x^3 + 6x - 20.)
It's conceivable that my solution below is essentially the only solution, as any reasonable solution will have to determine the minimal polynomial for the extension field that contains all of the roots for the *expression*.
I still hope that there is a better way to motivate the solution.
(%i1) eqn:x^3+3*c*x-20; 3 (%o1) x + 3 c x - 20 (%i2) root:rhs(solve(eqn,x)[3]); 3 1/3 c (%o2) (sqrt(c + 100) + 10) - ------------------------ 3 1/3 (sqrt(c + 100) + 10) (%i3) rootpick:pickapart(root,1); 3 1/3 (%t3) (sqrt(c + 100) + 10)
c (%t4) - ------------------------ 3 1/3 (sqrt(c + 100) + 10)
(%o4) %t4 + %t3 (%i5) %t5:%t4/(sqrt(c^3+100)-10)^(1/3),rootscontract; (%o5) - 1 (%i6) %t4:%t5*(sqrt(c^3+100)-10)^(1/3); 3 1/3 (%o6) - (sqrt(c + 100) - 10) (%i7) root:%t3+%t4; 3 1/3 3 1/3 (%o7) (sqrt(c + 100) + 10) - (sqrt(c + 100) - 10) (%i8) %,c=2; 3/2 1/3 3/2 1/3 (%o8) (2 3 + 10) - (2 3 - 10) (%i9) eqn; 3 (%o9) x + 3 c x - 20 (%i10) %,c=2,factor; 2 (%o10) (x - 2) (x + 2 x + 10)
At 10:43 AM 7/10/2020, Henry Baker wrote:
True.
Now do it like a high schooler...
At 09:17 AM 7/10/2020, Simon Plouffe wrote:
Hello,
the expression is simplified in a fraction of a second (sqrt(108)+10)^(1/3)-(sqrt(108)-10)^(1/3); to 2 by Maple.
Best regards,
Simon Plouffe
Le ven. 10 juil. 2020 à 17:47, Henry Baker <hbaker1@pipeline.com> a écrit :
(sqrt(108)+10)^(1/3)-(sqrt(108)-10)^(1/3)
[I thought this was a pretty cool problem that came up on the Maxima email list.]
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