Solve the following equation over the reals:$$ x^3 + 1 = 2\sqrt[3]{{2x - 1}} $$ I noticed that 1 is a trivial solution, then I tried raising the equation to the 3rd, then dividing the polynomial by $(x-1)$.. But I can't see the solution, how do I go from here?

  • $\begingroup$ See also this answer for some history behind this problem. $\endgroup$ – punctured dusk Feb 18 '16 at 10:34

We have $$x^3+1=2(2x-1)^{1/3}\iff x^3=2(2x-1)^{1/3} -1.$$ Here, setting $y=(2x-1)^{1/3}$ gives us $$y^3=2x-1 \ \ \text{and}\ \ x^3=2y-1.$$ Hence, we have $$\begin{align}y^3-x^3=(2x-1)-(2y-1)&\Rightarrow (y-x)(y^2+yx+x^2)=2(x-y)\\&\Rightarrow (y-x)(y^2+yx+x^2+2)=0\\&\Rightarrow (y-x)\{(x+(y/2))^2 + (3/4)y^2+2\}=0\\&\Rightarrow y=x.\end{align}$$ Hence, we have $$x^3=2x-1\iff (x-1)(x^2+x-1)=0\iff x=1,\frac{-1\pm\sqrt 5}{2}.$$


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?
  • 2
    $\begingroup$ For 2., rewrite it as $x^6+x^4+x^2(x+1)^2+2x^2+(x+1)^2+8$. $\endgroup$ – Micah Jul 21 '14 at 14:15
  • $\begingroup$ $x^2+x-1$ isn't nice factorable. $\endgroup$ – rae306 Jul 21 '14 at 14:17
  • $\begingroup$ 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 $\endgroup$ – Semiclassical Jul 21 '14 at 14:19
  • $\begingroup$ You actually came up with the 3 correct solutions @Semiclassical, those exactly 3 solutions were published soon after the exam was over. Thank you ! $\endgroup$ – Victor Jul 21 '14 at 18:32
  • 2
    $\begingroup$ 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. $\endgroup$ – Victor Aug 5 '14 at 11:23

$\ \frac{{x^3 + 1}}{2} = \sqrt[3]{{2x - 1}} \ $, denote LHS by $f(x)$ .

Since $f(x)$ is bijective, it must have an inverse function, particularly the RHS in this situation.So, the equation is simplified to $\ f(x) = f^{ - 1} (x) \ $, which further yields $\ f(f(x)) = x \ $. Since both functions are strictly increasing, it would only make sense that $f(x)=x$, therefore $\ x^3 + 1 - 2x = 0 \ $. $x=1$ is a trivial solution so we may rewrite our equation as $\ (x - 1)(x^2 + x - 1) = 0 \ $ .

By solving the quadratic we get $\ x_{1,2} = \frac{{ - 1 \pm \sqrt 5 }}{2} \ $ .We conclude that the real solutions of our equation are comprised by $\ S = \{ 1,\frac{{ - 1 \pm \sqrt 5 }}{2}\} \ $ .

  • $\begingroup$ In the general your solution is wrong. $f(x)=-x$ is also bijective. In our problem $f$ increases and now, your reasoning works. $\endgroup$ – Michael Rozenberg Jun 11 '19 at 15:53
  • $\begingroup$ I don't get your point, I specifically mentioned that I'm relying on the fact that $f$ increasing. $\endgroup$ – Victor Jun 12 '19 at 22:17
  • $\begingroup$ OK. Now I see it. $\endgroup$ – Michael Rozenberg Jun 13 '19 at 3:18

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 $$2\cdot y=\left(\frac{y^3+1}2\right)^3+1$$

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: $y-1 = 0 \implies y=1$

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

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

  • $\begingroup$ how do you plan on solving for $y$? $\endgroup$ – cirpis Jul 21 '14 at 14:06
  • 1
    $\begingroup$ $x = \frac{y^3+1}{2}$ $\endgroup$ – Daniel Fischer Jul 21 '14 at 14:06
  • $\begingroup$ As DanielFischer is pointing out, you've got a typo for $x=(y^3+1)/2$ $\endgroup$ – Semiclassical Jul 21 '14 at 14:11
  • $\begingroup$ 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. $\endgroup$ – Semiclassical Jul 21 '14 at 14:13
  • $\begingroup$ @Semiclassical, they are mapped to themselves. $\endgroup$ – Empy2 Feb 18 '16 at 11:24

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