First subtract 1 from both sides to get #x^3-2x-1 > 0#.
Notice that #x = -1# is a solution of #x^3-2x-1=0#, so #(x+1)# is a factor of #x^3-2x-1#. The other factor is #x^2-x-1#.
#x^3-2x-1=(x^2-x-1)(x+1)#.
Solve #x^2-x-1=0# using the quadratic formula
#x=(-b+-sqrt(b^2-4ac))/(2a)#
#=(1+-sqrt(1-4*1*(-1)))/(2*1)#
#=(1+-sqrt(5))/2#
So #x^3-2x-1# will cross the #x#-axis at #x = -1#, #x = (1-sqrt(5))/2# and #x=(1+sqrt(5))/2#.
Since the dominant term is #x^3#, the function #x^3-2x-1# will be negative for #x < -1#, positive for #-1 < x < (1-sqrt(5))/2#, negative for #(1-sqrt(5))/2 < x < (1+sqrt(5))/2# and positive for #x > (1+sqrt(5))/2#.
So the solution to the original problem is:
#-1 < x < (1-sqrt(5))/2# or #x > (1+sqrt(5))/2#.
