<s>[INST] How find this equation solution $2\sqrt[3]{2y-1}=y^3+1$ 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.
 [/INST] $$y^9+3y^6+3y^3-16y+9=0\iff(y-1)(y^2 + y - 1)(y^6 + 2y^4 + 2y^3 + 4y^2 + 2y +9) = 0$$
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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