Making $x$ the subject of $x^3-2x+y=1$ How to I make x the subject of this formula?
$$x^3-2x+y=1\tag1$$
My attempt:
I apply complete the square
$$(x-1)^3+y-1=1\tag2$$
$$(x-1)^3=2-y\tag3$$
$$x-1=\sqrt[3]{2-y}\tag4$$
$$x=1+\sqrt[3]{2-y}\tag5$$
but my teacher said it is wrong!
Can you please help? Thank in advance!

Use the formula for a cubic and find three solutions for $x$:


*

*$\frac{\sqrt[3]{2} \left(\sqrt{3} \sqrt{27 y^2-54 y-5}-9 y+9\right)^{2/3}+4 \sqrt[3]{3}}{6^{2/3} \sqrt[3]{\sqrt{3} \sqrt{27 y^2-54 y-5}-9 y+9}}$

*$\frac{i \sqrt[3]{2} \left(\sqrt{3}+i\right) \left(\sqrt{3} \sqrt{27 y^2-54 y-5}-9 y+9\right)^{2/3}-4 \sqrt[3]{3} \left(1+i \sqrt{3}\right)}{2\
   6^{2/3} \sqrt[3]{\sqrt{3} \sqrt{27 y^2-54 y-5}-9 y+9}}$

*$\frac{4 i \sqrt[3]{3} \left(\sqrt{3}+i\right)-\sqrt[3]{2} \left(1+i \sqrt{3}\right) \left(\sqrt{3} \sqrt{27 y^2-54 y-5}-9 y+9\right)^{2/3}}{2\
   6^{2/3} \sqrt[3]{\sqrt{3} \sqrt{27 y^2-54 y-5}-9 y+9}}$
