By the rational root theorem, any rational roots of #F(x) = 0# must be of the form #p/q# in lowest terms, where #p, q in ZZ#, #q != 0#, #p# a divisor of the constant term #15# and #q# a divisor of the coefficient #8# of the term #8x^3# of highest degree.
That means that the only possible rational roots are:
#+-15#, #+-15/2#, #+-5#, #+-15/4#, #+-3#, #+-5/2#, #+-15/8#, #+-3/2#, #+-5/4#, #+-1#, #+-3/4#, #+-5/8#, #+-1/2#, #+-3/8#, #+-1/4#, #+-1/8#
#F'(x) = 24x^2-6x+5 > 0# for all #x in RR#
So #F(x) = 0# has exactly one root.
#F(-1) = -8-3-5+15 = -1#
#F(-3/4) = -27/8-27/16-15/4+15#
#= (-54-27-60+240)/16 = 99/16#
So the root lies strictly between #-1# and #-3/4#
None of the possible rational roots lies in this range, so #F(x) = 0# has no rational roots.
graph{8x^3-3x^2+5x+15 [-1.688, 0.812, -0.62, 0.63]}
If you still want to solve the original problem, first let #t = x - 1/8#. This is called a Tschirnhaus transformation, giving us:
#F(x) = 8(t^3+37/64t+499/256)#
Next solve #t^3+37/64t+499/256 = 0# using Cardano's method
Let #t = u + v#
Then #0 = t^3+37/64t+499/256#
#=(u+v)^3+37/64(u+v)+499/256#
#=u^3+v^3+(3uv+37/64)(u+v)+499/256#
Now let #v = -37/(192u)# so that #3uv+37/64 = 0#
Then multiplying our equation through by #u^3# we get:
#(u^3)^2+499/256(u^3)-37^3/192^3 = 0#
Since this derivation is symmetric in #u# and #v# we find:
#t = root(3)((-499/256+sqrt((499/256)^2+4*37^3/192^3))/2) + root(3)((-499/256-sqrt((499/256)^2+4*37^3/192^3))/2)#
and
#x = 1/8 + root(3)((-499/256+sqrt((499/256)^2+4*37^3/192^3))/2) + root(3)((-499/256-sqrt((499/256)^2+4*37^3/192^3))/2)#
So it can be done, but it's a little complex.
