Solving the equation $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$ I wanna know how to solve this equation: $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$
I have some trouble to do that and I'd glad with any help I may get.

The algebraic $\frac{1}{12}(17 + \sqrt{97})$ is not a root of the equation
\begin{eqnarray}
-2 x^3 + 10 x^2 - 17 x + 8 = (2 x^2) (5 x - x^3)^{1/3}
\end{eqnarray}
Plugging it in, you find that the left hand side is real and equal to 
\begin{eqnarray}
\tfrac{1}{216}(-149 - 37 \sqrt{97}) = -2.37689 \dots
\end{eqnarray}
The right side is
\begin{eqnarray}
\tfrac{1}{432} (\tfrac{1}{2}( 595 - 61 \sqrt{97})^{1/3} (17 + \sqrt{97})^2 = 1.18844 \dots + i 2.05845 \dots
\end{eqnarray}
Note: $595 < 61 \sqrt{97}$. I think the ambiguity lies in the fact that we have not used the third-roots of unity. Numerical computations aside, just plot the two functions. The RHS is a positive function defined only in the I and II quadrants. The LHS is cubic. There is only one real intersection point.
