This method isn't for high schoolers, but should be doable for undergrads: Find the cube root of 10+sqrt(108). (This method works for all(?) fields Q(sqrt(D)), not just D=3.) 1. Find the minimal polynomial for 10+sqrt(108). (%i1) (x-(10+sqrt(108)))*(x-(10-sqrt(108))),expand; 2 (%o1) x - 20 x - 8 2. Produce the companion matrix for this minimal polynomial. (%i2) polytocompanion(%,x); [ 0 8 ] (%o2) [ ] [ 1 20 ] Its determinant is the 'norm' of 10+sqrt(108), which by inspection is -8. Note that -8 = (-2)^3; i.e., -8 is a perfect cube. The general 2x2 companion matrix looks like: [ 0 - Determinant ] (%o4) [ ] [ 1 Trace ] Let T=Trace and N=Determinant: (%i5) %,Trace=T,Determinant=N; [ 0 - N ] (%o5) [ ] [ 1 T ] Cube this matrix: (%i6) %^^3,factor; [ 2 ] [ - N T - N (T - N) ] (%o6) [ ] [ 2 2 ] [ T - N T (T - 2 N) ] 'Companionize' this matrix by producing the companion matrix with the same characteristic polynomial as our cubed matrix: (%i7) polytocompanion(charpoly(%,z),z),factor; [ 3 ] [ 0 - N ] (%o7) [ ] [ 2 ] [ 1 T (T - 3 N) ] Equate this matrix with our integer matrix above, and extract equations in N,T: (%i8) list_matrix_entries(col(%-matrix([0,8],[1,20]),2)); 3 2 (%o8) [- N - 8, T (T - 3 N) - 20] Eliminate N and factor: (%i9) eliminate(%,[N]),factor; 2 2 (%o9) [- (T - 2) (T + 2 T + 4) (T + 2 T + 10) 4 3 2 (T - 2 T - 6 T - 20 T + 100)] The simplest integer answer is T=2 and N=-2. So (10+sqrt(108))^(1/3) = (T/2) + sqrt((T/2)^2 - N) = 1 + sqrt( 1^2 - -2) = 1 + sqrt(3) 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.]