Proof that #N = (45+29 sqrt(2))^(1/3)+(45-29 sqrt(2))^(1/3)# is a integer ?

2 Answers
Sep 13, 2016

Answer:

Consider #t^3-21t-90 = 0#

This has one Real root which is #6# a.k.a. #(45+29sqrt(2))^(1/3)+(45-29sqrt(2))^(1/3)#

Explanation:

Consider the equation:

#t^3-21t-90 = 0#

Using Cardano's method to solve it, let #t = u+v#

Then:

#u^3+v^3+3(uv-7)(u+v)-90 = 0#

To eliminate the term in #(u+v)#, add the constraint #uv=7#

Then:

#u^3+7^3/u^3-90 = 0#

Multiply through by #u^3# and rearrange to get the quadratic in #u^3#:

#(u^3)^2-90(u^3)+343 = 0#

by the quadratic formula, this has roots:

#u^3 = (90+-sqrt(90^2-(4*343)))/2#

#color(white)(u^3) = 45 +- 1/2sqrt(8100-1372)#

#color(white)(u^3) = 45 +- 1/2sqrt(6728)#

#color(white)(u^3) = 45 +- 29sqrt(2)#

Since this is Real and the derivation was symmetric in #u# and #v#, we can use one of these roots for #u^3# and the other for #v^3# to deduce that the Real zero of #t^3-21t-90# is:

#t_1 = root(3)(45+29sqrt(2))+root(3)(45-29sqrt(2))#

but we find:

#(6)^3-21(6)-90 = 216 - 126 - 90 = 0#

So the Real zero of #t^3-21t-90# is #6#

So #6 = root(3)(45+29sqrt(2))+root(3)(45-29sqrt(2))#

#color(white)()#
Footnote

To find the cubic equation, I used Cardano's method backwards.

Sep 13, 2016

Answer:

#N = 6#

Explanation:

Making #x = 45+29 sqrt(2)# and #y = 45-29 sqrt(2)# then

#(x^(1/3)+y^(1/3))^3=x + 3 (x y)^(1/3)x^(1/3)+3(x y)^(1/3) y^(1/3)+y#

#(x y)^(1/3) = (7^3)^(1/3) = 7#
#x+y =2 xx 45#

so

#(x^(1/3)+y^(1/3))^3 = 90 + 21(x^(1/3)+y^(1/3))#

or calling #z = x^(1/3)+y^(1/3)# we have

#z^3-21 z-90 = 0#

with #90 = 2 xx 3^2 xx 5# and #z = 6# is a root so

#x^(1/3)+y^(1/3) = 6#