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This was one of the questions in my entry exam and I couldn't manage to solve it. I struggled with this equation for more than an hour, so I'd like a direct solution instead of hints please.

$$2\cdot (2x-1)^\frac13=x^3+1$$

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Were you supposed to solve for $x$? –  5xum Jul 21 '14 at 13:36
Well, $x=1$ is a solution. I arrived it at by just plugging it in. –  Mark Fantini Jul 21 '14 at 13:36
Yes, I was supposed to solve for x. –  Victor Jul 21 '14 at 13:36
x=1 is obvious, I need a way to reach the other/s. –  Victor Jul 21 '14 at 13:37
I think we need the exact wording from the entrance exam. Are we sure it did not say: "Find all integer solutions" ... or "How many real solutions are there" ?? –  GEdgar Jul 21 '14 at 14:42

2 Answers 2

up vote 4 down vote accepted

This solution will be frustratingly incomplete, since I'm not seeing an easy way to solve completely by hand. But here is where the road leads...

When solving equations in $x$, it is usually helpful to rewrite it in terms of a polynomial in $x$. To that end, cube both sides to obtain $8(2x-1)=(x^3+1)^3$, which upon moving terms to one side and expanding yields $$x^9+3x^6+3x^3-16x+9=0$$ Noting that $x=1$ is a trivial solution, we see that we can factorize this as $$(x-1)(x^8+x^7+x^6+4x^5+4x^4+4x^3+7x^2+7x-1)=0$$ One can in fact factor this further, though I should confess I only saw this after looking up the roots: $$(x-1)(x^2+x-1)(x^6+2x^4+2x^3+4x^2+2x+9)=0$$ The roots of the first two factors give three real roots which can be found by hand as $x=1,(-1\pm\sqrt{5})/2$. It turns out that the last factor has no real roots, and so these three are the result.

The two incomplete points:

  1. Does anyone know a good way to spot the factor of $x^2+x-1$?
  2. Does anyone see an obvious way to verify the lack of real roots in the last equation?
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For 2., rewrite it as $x^6+x^4+x^2(x+1)^2+2x^2+(x+1)^2+8$. –  Micah Jul 21 '14 at 14:15
$x^2+x-1$ isn't nice factorable. –  Rainier van Es Jul 21 '14 at 14:17
It's quadratic, so complete the square and solve. (Also, taking $x\to -x$ turns this into the equation for the golden ratio. So it's not an unfamiliar one.) @RainiervanEs –  Semiclassical Jul 21 '14 at 14:19
You actually came up with the 3 correct solutions @Semiclassical, those exactly 3 solutions were published soon after the exam was over. Thank you ! –  Victor Jul 21 '14 at 18:32
I've managed to come up with the following solution a few days ago : Given $$2\cdot (2x-1)^\frac13=x^3+1$$, we divide be 2 and denote the RHS as f(x). Further, we notice that the LHS is the inverse of this function and we get f(x)=f^(-1)(x). After proving that both function are monotonous we simplify the last equation to f(x)=x, which further gives us x^3+1-2x=0 . Since 1 is a trivial solution, we divide by (x-1), then take the quadratic solution, yielding the wished result. –  Victor Aug 5 '14 at 11:23

For $x$ - real $$2\cdot(2x-1)^\frac13=x^3+1$$

Let $y= (2x-1)^\frac13$

Therefore, $$y^3=2x-1$$ $$x=\frac{(y^3+1)}{2}$$

Then you get $$2\cdot y=(\frac{y^3+1}2)^3+1$$

Resolve it for $y$, and then replace find the $x$

Resolving for $y$:

  1. multiply both sides by 8 and you get

$$16\cdot y=(y^3+1)^3+8$$

$$16\cdot y = y^9+ 3\cdot y^6+3\cdot y^3 +9$$

$$(y-1)(y^6 + 2\cdot y^4+ 2\cdot y^3 +4\cdot y^2+2\cdot y+9 )(y^2 + y - 1)=0$$

Then we get: $y-1 = 0$ => $y=1$

$y^2 +y - 1=0$ => $y = \frac{1}{2}(-1\pm \sqrt5)$

$y^6 + 2\cdot y^4+ 2\cdot y^3 +4\cdot y^2+2\cdot y+9 =0$ => No roots

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how do you plan on solving for $y$? –  cirpis Jul 21 '14 at 14:06
$x = \frac{y^3+1}{2}$ –  Daniel Fischer Jul 21 '14 at 14:06
As DanielFischer is pointing out, you've got a typo for $x=(y^3+1)/2$ –  Semiclassical Jul 21 '14 at 14:11
And I'm skeptical of this approach: while it keeps 1 as a root, the other two irrational roots aren't mapped to anything nice. –  Semiclassical Jul 21 '14 at 14:13

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