\( \DeclareMathOperator{\abs}{abs} \newcommand{\ensuremath}[1]{\mbox{$#1$}} \)
(%i1) load(odes);
\[\tag{\%{}o1}\label{o1} /usr/share/maxima/5.38.1/share/contrib/odes/odes.mac\]
(%i2) eq: x^3 - 8*x^2 + x + 9 = 0;
\[\tag{eq}\label{eq}{{x}^{3}}-8{{x}^{2}}+x+9=0\]
(%i3) solvet(eq,x);
\[\tag{\%{}o3}\label{o3} [x=\frac{2\sqrt{61}\,\cos{\left( \frac{\operatorname{atan}\left( \frac{9\sqrt{5003}}{709}\right) }{3}\right) }+8}{3},x=\frac{2\sqrt{61}\,\cos{\left( \frac{\operatorname{atan}\left( \frac{9\sqrt{5003}}{709}\right) -2\ensuremath{\pi} }{3}\right) }+8}{3},x=\frac{2\sqrt{61}\,\cos{\left( \frac{\operatorname{atan}\left( \frac{9\sqrt{5003}}{709}\right) +2\ensuremath{\pi} }{3}\right) }+8}{3}]\]
(%i4) solve(eq,x);
\[\tag{\%{}o4}\label{o4} [x=\left( -\frac{\sqrt{3}\%{}i}{2}-\frac{1}{2}\right) \,{{\left( \frac{\sqrt{5003}\%{}i}{6}+\frac{709}{54}\right) }^{\frac{1}{3}}}+\frac{61\left( \frac{\sqrt{3}\%{}i}{2}-\frac{1}{2}\right) }{9{{\left( \frac{\sqrt{5003}\%{}i}{6}+\frac{709}{54}\right) }^{\frac{1}{3}}}}+\frac{8}{3},x=\left( \frac{\sqrt{3}\%{}i}{2}-\frac{1}{2}\right) \,{{\left( \frac{\sqrt{5003}\%{}i}{6}+\frac{709}{54}\right) }^{\frac{1}{3}}}+\frac{61\left( -\frac{\sqrt{3}\%{}i}{2}-\frac{1}{2}\right) }{9{{\left( \frac{\sqrt{5003}\%{}i}{6}+\frac{709}{54}\right) }^{\frac{1}{3}}}}+\frac{8}{3},x={{\left( \frac{\sqrt{5003}\%{}i}{6}+\frac{709}{54}\right) }^{\frac{1}{3}}}+\frac{61}{9{{\left( \frac{\sqrt{5003}\%{}i}{6}+\frac{709}{54}\right) }^{\frac{1}{3}}}}+\frac{8}{3}]\]

Attempting to verify, that the absolute of the cube root,
is actually the square root of (61/9)...

(%i5) expr : ((sqrt(5003)*%i)/6+709/54);
\[\tag{expr}\label{expr}\frac{\sqrt{5003}\%{}i}{6}+\frac{709}{54}\]
(%i6) abs(expr)^(1/3);
\[\tag{\%{}o6}\label{o6} \frac{\sqrt{61}}{3}\]

Because this time, the base was real, its cube root
was also treated as real.

(%i7) (61/9)/(%o6);
\[\tag{\%{}o7}\label{o7} \frac{\sqrt{61}}{3}\]

Therefore, two terms are being added, that have the same absolute.

Is Maxima able to give me the numbers from both forms?

(%i8) float(%o3);
\[\tag{\%{}o8}\label{o8} [x=7.719423315394779,x=1.229127641686919,x=-0.9485509570816971]\]
(%i9) float(%o4);
\[\tag{\%{}o9}\label{o9} [x=\left( -0.8660254037844386\%{}i-0.5\right) \,{{\left( 11.78864802351067\%{}i+13.12962962962963\right) }^{\frac{1}{3}}}+\frac{6.777777777777778\left( 0.8660254037844386\%{}i-0.5\right) }{{{\left( 11.78864802351067\%{}i+13.12962962962963\right) }^{\frac{1}{3}}}}+2.666666666666666,x=\left( 0.8660254037844386\%{}i-0.5\right) \,{{\left( 11.78864802351067\%{}i+13.12962962962963\right) }^{\frac{1}{3}}}+\frac{6.777777777777778\left( -0.8660254037844386\%{}i-0.5\right) }{{{\left( 11.78864802351067\%{}i+13.12962962962963\right) }^{\frac{1}{3}}}}+2.666666666666666,x={{\left( 11.78864802351067\%{}i+13.12962962962963\right) }^{\frac{1}{3}}}+\frac{6.777777777777778}{{{\left( 11.78864802351067\%{}i+13.12962962962963\right) }^{\frac{1}{3}}}}+2.666666666666666]\]

Conclusion:
If Maxima is unable to compute the numeric
version of one solution set, how is the
average user expected to do so? And yet, how
is a set of exact, non-numeric roots to become
useful, if their approximate, numeric values
cannot be verified?

(%i10) s : map(rhs,%o8);
\[\tag{s}\label{s}[7.719423315394779,1.229127641686919,-0.9485509570816971]\]
(%i11) ProductSeries(list) := block (
   product : 1,
   for elem in list do (
       product : product * -elem
   ),
   product
)$
(%i12) ProductSeries(s);
\[\tag{\%{}o12}\label{o12} 8.99999999999998\]

The constant term was (+9)...

The following Maxima function does, what my own suggested
program also does: Compute numeric approximations only:

(%i13) allroots(eq);
\[\tag{\%{}o13}\label{o13} [x=1.229127641686918,x=-0.9485509570816991,x=7.71942331539478]\]
(%i14) s : map(rhs,%o13)$
(%i15) ProductSeries(s);
\[\tag{\%{}o15}\label{o15} 8.999999999999997\]
Created with wxMaxima.