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.]