# Tough equation in my entry exam

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

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. –  rae306 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