What is the root of #x^3 - x - 1 = 0# in exact form?

Question

Elias Ackerman · Accepted Answer

Real root:

#root(3)((27+3sqrt(69))/54)+root(3)((27-3sqrt(69))/54)#

and related complex roots...

Given: #x^3-x-1 = 0# Using Cardano's method, let #x = u+v# Then: #u^3+v^3+(3uv-1)(u+v)-1 = 0# To eliminate the term in #(u+v)# add the constraint: #3uv-1 = 0# So: #v = 1/(3u)# and our equation can be rewritten: #u^3+1/(27u^3)-1 = 0# Multiply through by #27u^3# and rearrange a little to get: #27(u^3)^2-27(u^3)+1 = 0# Hence by the quadratic formula: #u^3 = (27+-sqrt(27^2-4(27)(1)))/(2*27)# #color(white)(u^3) = (27+-sqrt(729-108))/54# #color(white)(u^3) = (27+-sqrt(621))/54# #color(white)(u^3) = (27+-3sqrt(69))/54# Since these are both real values and the derivation is symmetric in #u# and #v#, we can use one for #u^3# and the other for #v^3# to find real root of our original cubic: #x_1 = root(3)((27+3sqrt(69))/54)+root(3)((27-3sqrt(69))/54)# and related complex roots: #x_2 = omega root(3)((27+3sqrt(69))/54)+omega^2 root(3)((27-3sqrt(69))/54)# #x_3 = omega^2 root(3)((27+3sqrt(69))/54)+omega root(3)((27-3sqrt(69))/54)# where #omega = -1/2+sqrt(3)/2i# is the primitive complex cube root of #1#.