Re: [math-fun] High schoolers: simplify this expression
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.]
SymPy's main `simplify` doesn't handle this but the procedure `minpoly` (minimal polynomial having an algebraic element as a root) can be used:
from sympy import *
simplify((sqrt(108)+10)**(Integer(1)/3)-(sqrt(108)-10)**(Integer(1)/3)) # not effective 2**(1/3)*(-(-5 + 3*sqrt(3))**(1/3) + (5 + 3*sqrt(3))**(1/3))
minpoly(_) x - 2
Adding a minpoly based simplification to the general simplify procedure recently came up on their issue tracker: https://github.com/sympy/sympy/issues/19726 (there look to be open issues suggesting a more general approach using the routine `ratsimpmodprime`). (not a high school answer when I attended) On Fri, Jul 10, 2020 at 1:44 PM Henry Baker <hbaker1@pipeline.com> 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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On 10/07/2020 18:43, Henry Baker wrote:
True.
Now do it like a high schooler...
Step 1: either use a calculator or do some rough pencil-and-paper approximation to guess that the answer might be 2. Step 2: play around with those two square-rooted quantities a bit, finding that their product is the cube root of 8, namely 2 [citation needed]. Step 3: if we call the one with the + sign "A" then this means we're trying to prove that A - 2/A = 2. Step 4: Wait, we know what the solutions of that are; they're [pulls out quadratic formula] 1 +- sqrt(3). So that ought to mean that A is 1 + sqrt(3). Step 5: Waaait, we can check that easily. What's the _cube_ of 1 + sqrt(3)? [scribbles] 10 + 6 sqrt(3), which is indeed the same thing as sqrt(108) + 10. Step 6: OK, so we've just proved that our first quantity is 1 + sqrt(3). So is the second one sqrt(3) - 1? That'll be true if _that_ cubes up to sqrt(108)-10. [scribbles] Yes, it does. (For extra points, "Of course it does" without the scribbling, but that might be a bit much to expect of a high-schooler.) And we're done. Nothing there should be beyond the wit of, say, a bright 15-year-old. -- g
* Henry Baker <hbaker1@pipeline.com> [Jul 11. 2020 16:49]:
True.
Now do it like a high schooler...
Search the internet, you mean? And I wish this was true only for high schoolers. The concept of "book" (or, god forbid, "journal article") appears to be disappearing young people. Best regards, jj
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. [...]
Hey --- lighten up, gramps! WFL On 7/11/20, Joerg Arndt <arndt@jjj.de> wrote:
* Henry Baker <hbaker1@pipeline.com> [Jul 11. 2020 16:49]:
True.
Now do it like a high schooler...
Search the internet, you mean? And I wish this was true only for high schoolers. The concept of "book" (or, god forbid, "journal article") appears to be disappearing young people.
Best regards, jj
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. [...]
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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? 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.]
Here's another way to do it (like Gareth's approach), which is at high school level, but sort of requires that you guess part of the solution in advance: (spoiler space) . . . . . . . . . . . Another way is to hypothesize that the expressions (sqrt(108) + 10)^(1/3) and (sqrt(108) - 10)^(1/3) might each be equal to a + b sqrt(3) for some a and b, since sqrt(108) = 6 sqrt(3). Cubing both sides gives +- 10 + 6 sqrt(3) == a (a^2 + 9 b^2) + 3 b (a^2 + b^2) sqrt(3) or equating coefficients: +- 10 == a (a^2 + 9 b^2) 2 == b (a^2 + b^2) I think the problem with general coefficients on the left-hand side is as hard as simplifying the original cubic, but if you guess that a and b are integers, then you quickly get a = 1 and b = 1 for the +10 case and a = -1 and b = 1 for the -10 case, i.e. (sqrt(108) + 10)^(1/3) = 1 + sqrt(3) (sqrt(108) - 10)^(1/3) = -1 + sqrt(3) so the answer is (1 + sqrt(3)) - (-1 + sqrt(3)) = 2. (I can't remember the book I saw this in originally; I think Gareth's approach also seems a bit less brute-force.) --Neil
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.]
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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.]
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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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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participants (9)
-
Ehren Metcalfe -
Fred Lunnon -
Gareth McCaughan -
Henry Baker -
Joerg Arndt -
Michael Greenwald -
Neil Bickford -
rcs＠xmission.com -
Simon Plouffe