1

If $x\in \mathbb{R}$, and $$x ^3+1 =2\sqrt [ 3 ]{ 2x-1 }$$ find $x$.

I asked my teacher to help me in solving it, he responded I can't! So, I hope you help me in solving it or even giving me hints that will help me in solving.

user26857
  • 52,094
Math Lover
  • 738
  • 4
  • 12

4 Answers4

3

This is one of the so-called Jewish Problems. Its solution can be found here (problem 4): http://arxiv.org/pdf/1110.1556v2.pdf. There's an interesting story behind them; let me cite a few sentences:

I was told that these problems were carefully designed to have elementary solutions (so that the Department could avoid scandals) that were nearly impossible to find. Any student who failed to answer could easily be rejected, so this system was an effective method of controlling admissions.

Here's another way to write up the solution in the above paper.

Let $g(x)=\frac{x^3+1}2$; we want to find the fixed points of $g\circ g$. $g$ is strictly increasing, so if $g(x)\overset<>x$ then $g(g(x))\overset<>g(x)\overset<>x$. Hence every fixed point should satisfy $g(x)=x$, i.e. $$x^3-2x+1=0.$$

$x=1$ is one solution; factoring out $x-1$ gives $x=\frac{-1\pm\sqrt5}2$.

Bart Michels
  • 26,355
2

Cubing we get $x^9+3x^6+3x^3+1=16x-8$ thus simplifying we get $x^9+3x^6+3x^3-16x+9=0$ so the obvious real solution is $1$ others can be found out but i think its very complex.

Archis Welankar
  • 15,910
2

By factoring we see that $$ x^9+3x^6+3x^3-16x+9=(x^6 + 2x^4 + 2x^3 + 4x^2 + 2x + 9)(x^2 + x - 1)(x - 1), $$ so that we have solutions $x=1$, $x=\frac{\pm \sqrt{5}-1}{2}$ and $6$ further complex solutions. So we have exactly three real solutions, easy to compute.

Dietrich Burde
  • 130,978
  • The degree-six equals $(x^3+1)^2+(2x^2+1)^2/2+(2x+1)^2/2+7$, to confirm the other six solutions are complex. But are all the roots of the degree-9 equation roots of the original equation? – Empy2 Feb 18 '16 at 10:26
  • We only have to check it for the three real solutions, or is the OP considering all complex solutions ? – Dietrich Burde Feb 18 '16 at 10:30
  • Even for the real negative one, i.e., $-(1+\sqrt{5})/2$, it depends on which value of $\sqrt[3]{z}$ do you use when $z$ is negative. I remember that in school often they write things $\sqrt[3]{-8} =-2$ but it's a convention. – Giovanni Resta Feb 18 '16 at 10:31
1

We write $(x^3+1)^3=8(2x-1)$, which simplifies into $$x^9+3 x^6+3 x^3-16 x+9 = 0$$

With a little messing around, we see that $x\in\{1,-\phi,\phi-1\}$ are solutions (where $\phi$ denotes the golden ratio). Thus, we may divide the polynomial by $(x-1)(x+\phi)(x-\phi+1)=x^3-2x+1$. We get $$x^6 + 2 x^4 + 2 x^3 + 4 x^2 + 2 x + 9=0$$ This however is a sixth degree polynomial and so it cannot be solved algebraically in terms of a finite number of additions, subtractions, multiplications, divisions, and root extractions. In other words, the other 6 (complex) solutions can most likely not be expressed in any "nice way".