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.
 A: Cube both sides and collect terms. You should get
\begin{eqnarray}
512 - 3264 x + 8856 x^2 - 13457 x^3 + 12702 x^4 - 7794 x^5 + 
 3136 x^6 - 844 x^7 + 120 x^8 = 0
\end{eqnarray}
which factorizes into
\begin{eqnarray}
(8 - 17 x + 6 x^2) (64 - 272 x + 481 x^2 - 456 x^3 + 258 x^4 - 
   84 x^5 + 20 x^6) = 0.
\end{eqnarray}
Since this is an $8^{\text{th}}$-degree polynomial equation, you can use Mathematica or a calculator to determine six of the eight solutions (by solving the $6^{\text{th}}$-degree polynomial equation numerically) and the other two by solving the quadratic equation: $\frac{1}{12}(17 \pm \sqrt{97})$. The 6 solutions are 3 pairs of complex conjugates:
\begin{eqnarray}
& & 0.647522 \dots \pm i 2.21209 \dots  \
& & 0.657656 \dots \pm i 0.218497 \dots  \
& & 0.794822 \dots \pm i 0.788971 \dots 
\end{eqnarray}
Of course, you'll have to check that these solutions work. At most one real solution and two complex solutions are spurious.
A: Note that $\frac{1}{12} \left(17+\sqrt{97}\right)$ is not a solution. It just appeared because of cubing.
A: Once someone gives the answer, it becomes easier!
I presume you are only looking for real roots and that $\displaystyle (5x-x^3)^{1/3}$ is the unique real number whose cube is $\displaystyle 5x - x^3$.
First observe that
$\displaystyle (x-2)^3 + 5x - x^3 = -6x^2 + 17x -8$
So if $\displaystyle a = (5x-x^3)^{1/3}$ and $\displaystyle b = x-2$
then the equation becomes
$$\displaystyle -2x^2(x-2) - (-6x^2 - 17x - 8) = 2x^2(5x-x^3)^{1/3}$$
and thus
$$\displaystyle -2x^2 b - (a^3 + b^3) = 2x^2a$$
and so
$$\displaystyle (a+b)(a^2 - ab + b^2 + 2x^2) = 0$$
Now for any real numbers $\displaystyle a,b,x$ we have that
$\displaystyle a^2 -ab + b^2 \ge 0$ and  $\displaystyle 2x^2 \ge 0$ and thus we must have that
$\displaystyle a + b = 0$ or $\displaystyle 2x^2 = 0$
$\displaystyle a = -b$ can be cubed to give $\displaystyle 6x^2 - 17x + 8 = 0$.
$\displaystyle 2x^2 = 0$ can be easily eliminated.
Note that the transformations we did were equivalent, and so both roots of $\displaystyle 6x^2 - 17x + 8 = 0$ are also roots of the original equation, given the definition of cuberoot at the top of the answer.
If you define $\displaystyle z^{1/3}$ using the principal branch of $\log z$, then the above assumption of $\displaystyle a$ being real is valid only if $\displaystyle 5x - x^3 \ge 0$, which eliminates $\displaystyle \dfrac{17 + \sqrt{97}}{12}$ as a root.
A: 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.
