$$\DeclareMathOperator{\abs}{abs} \newcommand{\ensuremath}[1]{\mbox{#1}}$$
$\tag{\%{}o1}\label{o1} /usr/share/maxima/5.38.1/share/contrib/odes/odes.mac$
 (%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.