Solving an equation over the reals: $ x^3 + 1 = 2\sqrt{{2x – 1}}$

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?

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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?

$\
\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}\}
\
$ .

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