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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?

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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: $x=1$.

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  • $\begingroup$ So it means that this quadratic equation will have three roots? $\endgroup$ – S.Bansal Jan 18 '18 at 13:14
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    $\begingroup$ So it has three roots $\endgroup$ – S.Bansal Jan 18 '18 at 13:16
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    $\begingroup$ I think it's not quadratic equation. $\endgroup$ – Michael Rozenberg Jan 18 '18 at 13:16
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    $\begingroup$ After substitution $t=\sqrt[3]x$ it's a cubic equation. $\endgroup$ – Michael Rozenberg Jan 18 '18 at 13:19
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    $\begingroup$ You are welcome! $\endgroup$ – Michael Rozenberg Jan 24 '18 at 9:22
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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 $x$.

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  • $\begingroup$ I like the answer format of including a spoiler! $\endgroup$ – Brian J Jan 18 '18 at 18:39
  • $\begingroup$ Ah yes, of course. Thanks @PaulSinclair $\endgroup$ – Dave Jan 18 '18 at 19:08
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write $$\sqrt[3]{x}=2-x$$ and raise to the power three your equation is equivalent to $$-(x-1) \left(x^2-5 x+8\right)=0$$

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