Hello, I posted the '2' answer for a good reason, here is a very simple equation : x^4 - x^3 +1. The solutions have a resemblance to the equation posted earlier. If you can find this solution then you are very good. Here is the solution lprinted, it is horrible. There is a method for solving this of course. Honestly, I never met anyone that could write down this solution without using Maple or mathematica. I often show this equation to my students. Solution of x^4 - x^3 +1 : 1/4+1/12*3^(1/2)*((2*(108+12*I*687^(1/2))^(2/3)+3*(108+12*I*687^(1/2))^(1/3)+96 )/(108+12*I*687^(1/2))^(1/3))^(1/2)+1/12*(-6*((108+12*I*687^(1/2))^(2/3)*((2*( 108+12*I*687^(1/2))^(2/3)+3*(108+12*I*687^(1/2))^(1/3)+96)/(108+12*I*687^(1/2)) ^(1/3))^(1/2)-3*3^(1/2)*(108+12*I*687^(1/2))^(1/3)-3*(108+12*I*687^(1/2))^(1/3) *((2*(108+12*I*687^(1/2))^(2/3)+3*(108+12*I*687^(1/2))^(1/3)+96)/(108+12*I*687^ (1/2))^(1/3))^(1/2)+48*((2*(108+12*I*687^(1/2))^(2/3)+3*(108+12*I*687^(1/2))^(1 /3)+96)/(108+12*I*687^(1/2))^(1/3))^(1/2))/(108+12*I*687^(1/2))^(1/3)/((2*(108+ 12*I*687^(1/2))^(2/3)+3*(108+12*I*687^(1/2))^(1/3)+96)/(108+12*I*687^(1/2))^(1/ 3))^(1/2))^(1/2), 1/4+1/12*3^(1/2)*((2*(108+12*I*687^(1/2))^(2/3)+3*(108+12*I* 687^(1/2))^(1/3)+96)/(108+12*I*687^(1/2))^(1/3))^(1/2)-1/12*(-6*((108+12*I*687^ (1/2))^(2/3)*((2*(108+12*I*687^(1/2))^(2/3)+3*(108+12*I*687^(1/2))^(1/3)+96)/( 108+12*I*687^(1/2))^(1/3))^(1/2)-3*3^(1/2)*(108+12*I*687^(1/2))^(1/3)-3*(108+12 *I*687^(1/2))^(1/3)*((2*(108+12*I*687^(1/2))^(2/3)+3*(108+12*I*687^(1/2))^(1/3) +96)/(108+12*I*687^(1/2))^(1/3))^(1/2)+48*((2*(108+12*I*687^(1/2))^(2/3)+3*(108 +12*I*687^(1/2))^(1/3)+96)/(108+12*I*687^(1/2))^(1/3))^(1/2))/(108+12*I*687^(1/ 2))^(1/3)/((2*(108+12*I*687^(1/2))^(2/3)+3*(108+12*I*687^(1/2))^(1/3)+96)/(108+ 12*I*687^(1/2))^(1/3))^(1/2))^(1/2), 1/4-1/12*3^(1/2)*((2*(108+12*I*687^(1/2))^ (2/3)+3*(108+12*I*687^(1/2))^(1/3)+96)/(108+12*I*687^(1/2))^(1/3))^(1/2)+1/12*( -6*((108+12*I*687^(1/2))^(2/3)*((2*(108+12*I*687^(1/2))^(2/3)+3*(108+12*I*687^( 1/2))^(1/3)+96)/(108+12*I*687^(1/2))^(1/3))^(1/2)+3*3^(1/2)*(108+12*I*687^(1/2) )^(1/3)-3*(108+12*I*687^(1/2))^(1/3)*((2*(108+12*I*687^(1/2))^(2/3)+3*(108+12*I *687^(1/2))^(1/3)+96)/(108+12*I*687^(1/2))^(1/3))^(1/2)+48*((2*(108+12*I*687^(1 /2))^(2/3)+3*(108+12*I*687^(1/2))^(1/3)+96)/(108+12*I*687^(1/2))^(1/3))^(1/2))/ (108+12*I*687^(1/2))^(1/3)/((2*(108+12*I*687^(1/2))^(2/3)+3*(108+12*I*687^(1/2) )^(1/3)+96)/(108+12*I*687^(1/2))^(1/3))^(1/2))^(1/2), 1/4-1/12*3^(1/2)*((2*(108 +12*I*687^(1/2))^(2/3)+3*(108+12*I*687^(1/2))^(1/3)+96)/(108+12*I*687^(1/2))^(1 /3))^(1/2)-1/12*(-6*((108+12*I*687^(1/2))^(2/3)*((2*(108+12*I*687^(1/2))^(2/3)+ 3*(108+12*I*687^(1/2))^(1/3)+96)/(108+12*I*687^(1/2))^(1/3))^(1/2)+3*3^(1/2)*( 108+12*I*687^(1/2))^(1/3)-3*(108+12*I*687^(1/2))^(1/3)*((2*(108+12*I*687^(1/2)) ^(2/3)+3*(108+12*I*687^(1/2))^(1/3)+96)/(108+12*I*687^(1/2))^(1/3))^(1/2)+48*(( 2*(108+12*I*687^(1/2))^(2/3)+3*(108+12*I*687^(1/2))^(1/3)+96)/(108+12*I*687^(1/ 2))^(1/3))^(1/2))/(108+12*I*687^(1/2))^(1/3)/((2*(108+12*I*687^(1/2))^(2/3)+3*( 108+12*I*687^(1/2))^(1/3)+96)/(108+12*I*687^(1/2))^(1/3))^(1/2))^(1/2) Here is the first solution : there are 4 of them, very similar : 1/4+1/12*3^(1/2)*((2*(108+12*I*687^(1/2))^(2/3)+3*(108+12*I*687^(1/2))^(1/3)+96 )/(108+12*I*687^(1/2))^(1/3))^(1/2)+1/12*(-6*((108+12*I*687^(1/2))^(2/3)*((2*( 108+12*I*687^(1/2))^(2/3)+3*(108+12*I*687^(1/2))^(1/3)+96)/(108+12*I*687^(1/2)) ^(1/3))^(1/2)-3*3^(1/2)*(108+12*I*687^(1/2))^(1/3)-3*(108+12*I*687^(1/2))^(1/3) *((2*(108+12*I*687^(1/2))^(2/3)+3*(108+12*I*687^(1/2))^(1/3)+96)/(108+12*I*687^ (1/2))^(1/3))^(1/2)+48*((2*(108+12*I*687^(1/2))^(2/3)+3*(108+12*I*687^(1/2))^(1 /3)+96)/(108+12*I*687^(1/2))^(1/3))^(1/2))/(108+12*I*687^(1/2))^(1/3)/((2*(108+ 12*I*687^(1/2))^(2/3)+3*(108+12*I*687^(1/2))^(1/3)+96)/(108+12*I*687^(1/2))^(1/ 3))^(1/2))^(1/2) Have fun, ! Simon Plouffe Le dim. 12 juil. 2020 à 19:36, <rcs@xmission.com> a écrit :
I'll quibble with "every solution".
The sqrt(108) is begging to be changed into 6 sqrt3.
Looking at 10 + 6 sqrt3, its worth trying the guess A + B sqrt3 for the cube root, and expanding into
10 + 6 sqrt3 = A^3 + 3 A^2 B sqrt3 + 3 A B^2 3 + B^3 3 sqrt3.
Then separate out the stuff multiplied by sqrt3, giving
6 sqrt3 = 3 A^2 B sqrt3 + 3 B^3 sqrt3,
and the leftover 10 = A^3 + 9 A B^2.
Hoping for an integer solution, we notice A is a divisor of 10, and B divides 2. A couple of trials finds A and B.
Now looking at the other cube root term, since the pieces are the same as before, with some sign changes, the next guess is that some version of +-A and +-B will work for the second cube root.
No need to learn about polynomials, the factor theorem, etc.
Rich
----- Quoting Henry Baker <hbaker1@pipeline.com>:
Yes, I agree with you, Michael.
Your solution also seems to be what I *might* have done in H.S., were I sufficiently smart.
I don't think I would have done it within the confines of a math A.P. test, however, even with the help of a symbolic calculator (they were just then being invented!).
In any case, your solution confirms my intuition, which is that *any/every* solution would -- in effect -- have to construct the minimal polynomial and factor it, although I didn't know anything about minimal polynomials in H.S.
At 10:51 PM 7/11/2020, Michael Greenwald wrote:
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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