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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.

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    I´m preparing to a proof that has high reputation, then i collect questions in forums which have this purpose, that was the way i found out that exercise.In addition, the proof is high school level. I´m not sure that equation has some solution based in that grade of knowledge. Sorry about my english.2010-11-04

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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.

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    1/3 is a exponent2010-11-04
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Note that $\frac{1}{12} \left(17+\sqrt{97}\right)$ is not a solution. It just appeared because of cubing.

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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.

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    @Isaac: Thanks, I will edit the answer to remove that and just mention principal branch of log z.2010-12-08
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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.

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    @J.M: Yes, I gathered that (and hence changed my answer to reflect that) :-)2010-11-05