Polynomial Equation Problem with Complex Roots

Question

How could one find the complex or imaginary roots of any equations like

$$x+\sqrt[3]{x}-2=0$$

One of its roots is $1$, but what is/are the other/s?

Answer 1

Let $\sqrt[3]{x}=t$.

Hence, we have: $$t^3+t-2=0$$ or: $$t^3-t^2+t^2-t+2t-2=0$$ or: $$(t-1)(t^2+t+2)=0,$$ which gives: $$\sqrt[3]{x}=-\frac{1}{2}\pm\frac{\sqrt7}{2}i$$ or: $x=1$.

Answer 2

write $$\sqrt[3]{x}=2-x$$ and raise to the power three your equation is equivalent to $$-(x-1) \left(x^2-5 x+8\right)=0$$

Answer 3

One can make the substitution $t^3=x$ so that the equation becomes a cubic in $t$: $$t^3+t-2=0$$ As you say, $x=1\iff t=1$ is a solution, so we can divide the polynomial in $t$ by $t-1$ to obtain a quadratic (I leave the long division or factoring to you). Then the resulting quadratic can be solved by the quadratic formula (or factoring if its easy enough).

So you can check your long division: the resulting quadratic should be $$t^2+t+2$$, and this has negative discriminant so it has complex roots.

Then you just use $x=t^3$ to obtain the solutions in $x$.