# 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.

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I am curious. How did you come across this equation? –  Aryabhata Nov 4 '10 at 22:38
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. –  tom Nov 4 '10 at 23:07
Which forum did you find it in? Perhaps a link? –  Aryabhata Nov 4 '10 at 23:08
@tom: I'd like the link - any time I can practice my portuguese is good for me! –  Jason DeVito Nov 4 '10 at 23:23
The link of the community is orkut.com.br/Main#Community?cmm=1299345 –  tom Nov 5 '10 at 4:31

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

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I hate it when downvoters don't explain themselves... they're being jerks. –  Ｊ. Ｍ. Nov 4 '10 at 22:27
I imagine the claim is this should be a comment. But I will upvote to counteract it. The statement is quite useful. –  Ross Millikan Nov 4 '10 at 22:35
thanks, it was a important note. –  tom Nov 4 '10 at 23:29
Are you sure? If x and y are real, then x^3 = y^3 implies x = y. –  Aryabhata Nov 5 '10 at 1:07
@J.M, Ross: Maybe the downvoters did this because the statement is false? btw, I don't see any downvotes! –  Aryabhata Nov 5 '10 at 1:08

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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Really? Is it impossible solve it by factorize or trick of algebra? Sorry, I´m just a high school grade studant. –  tom Nov 4 '10 at 21:12
If I'm reading your equation correctly, no, not if you are seeking all solutions. Is the 1/3 an exponent? Please use latex formatting. It's tough to tell what you mean. –  user02138 Nov 4 '10 at 21:14
1/3 is a exponent –  tom Nov 4 '10 at 23:09
I am not sure if "at least one real solution is spurious" is correct. Did you actually verify? –  Aryabhata Nov 5 '10 at 1:14
Checking Wolfram Alpha (wolframalpha.com/input/…), the three valid solutions are $\frac{1}{{12}}\left( {17 - \sqrt {97} } \right)$ and $0.64752 \pm 2.21209i$. –  Alexsander Akers Nov 5 '10 at 2:31

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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I´d only like to know how did you imagine the factorize of begining - (w-2)^3 +5w - w^3=-6w^2 +17w -8 - did you look to second member of equation? That was wonderful. Thanks very much. –  tom Nov 4 '10 at 23:49
@Tom: Knowing that is was a factor (based on user02138's solution) helped. I am not sure exactly how I came up with it. Sorry. –  Aryabhata Nov 5 '10 at 1:12
it make sense, I appreciate your attention. –  tom Nov 5 '10 at 1:21
"If you define $z^{1/3}$ as the complex number with the least argument (which has become pretty standard convention nowadays)"—depends on exactly what you meant by least argument. I have seen $z^{1/3}$ defined as having argument in either $[0,\frac{2\pi}{3})$ or argument in $(-\frac{\pi}{3},\frac{\pi}{3}]$, the latter being more common in technological implementations. –  Isaac Dec 8 '10 at 8:03
@Isaac: Thanks, I will edit the answer to remove that and just mention principal branch of log z. –  Aryabhata Dec 8 '10 at 8:09

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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cuberoot of an arbitrary real R number is well defined: the unique real root of x^3 = R. In this case 5x-x^3 becomes negative and your software seems to be getting unnecessarily "complex". –  Aryabhata Nov 5 '10 at 3:38
It depends on the definition of (5x-x^3)^(1/3). Given that this is a high school level problem, I presume we are only talking of the real values of the three possible values. I am not sure what is standard, though. Apparently, Mathematica seems to take the one with positive imaginary part. –  Aryabhata Nov 5 '10 at 5:35
@Moron: The approach taken in Mathematica is to take the root that is consistent with the principal value of the logarithm, which is the same as the conventional cube root when the argument is positive. That's one reason there is an add-on package in Mathematica that forces the cube root to be $\mathrm{sgn}(x)\sqrt[3]{|x|}$ to agree with the "usual" cube root, to cater for less-advanced users. –  Ｊ. Ｍ. Nov 5 '10 at 11:05
@J.M: Yes, I gathered that (and hence changed my answer to reflect that) :-) –  Aryabhata Nov 5 '10 at 13:16