How find this equation solution $2\sqrt[3]{2y-1}=y^3+1$

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find this equation roots: $$2\sqrt[3]{2y-1}=y^3+1$$

My try: since $$8(2y-1)=(y^3+1)^3=y^9+1+3y^3(y^3+1)$$ then $$y^9+3y^6+3y^3-16y+9=0$$

Then I can't.Thank you someone can take hand find the equation roots.

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3 Answers

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Let $f(y) = \frac12 (y^3+1)$, we have

$$2\sqrt[3]{2y-1} = y^3+1\quad\iff\quad f^{-1}(y) = f(y) \quad\implies\quad y = f(f(y))$$

Since $f(y)$ is a strictly increasing function in $y$, we have

• If $f(x) > x$, then $f(f(x)) > x$.
• If $f(x) < x$, then $f(f(x)) < x$.

This means

$$\begin{align} y = f(f(y)) \implies & y = f(y)\\ \iff & y^3+1 - 2y = 0\\ \iff & (y^2 + y -1 )(y-1) = 0\\ \implies & y = 1\text{ or } \frac{-1\pm \sqrt{5}}{2} \end{align}$$

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Another approach:

Put $x=\sqrt[3]{2y-1}$ then $x^3=2y-1$. So, $x^3+1=2y$

Then we have the system of equation:

$\left\{\begin{matrix} x^3+1=2y\\y^3+1=2x \end{matrix}\right.$

Subtracting these two equations from each other, we have

$(x-y)(x^2+xy+y^2+2)=0$ which gives $x=y$ as ($x^2+xy+y^2+2>0 $ $\forall$ $x,y$)

Hence we have $y=\sqrt[3]{2y-1}$

$\Rightarrow y^3-2y+1=0$

$\Leftrightarrow(y-1)(y^2+y-1)=0$

Therefore, $y=1$ or $y=\frac{-1\pm \sqrt{5} }{2}$

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$$y^9+3y^6+3y^3-16y+9=0\iff(y-1)(y^2 + y - 1)(y^6 + 2y^4 + 2y^3 + 4y^2 + 2y +9) = 0$$

Three real-valued roots: $$y = 1, y = \dfrac{-1 \pm \sqrt 5}{2}$$

I noticed by inspection that $y = 1$ solves the equation. Then, using polynomial division, and manipulation with the resulting quotient, was able to find the quadratic factor. The sixth degree polynomial has no real roots.

Additional tip: Sometimes it helps to graph the equation to learn how many real roots there are: look for points intersecting the x-axis. Here's the graph of your 9th-degree polynomial:

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