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The equation $$\sqrt[3]{4-x^2}+\sqrt{x^2-3}=1$$ has a solution $x=\pm2\sqrt{3}$ but

Solve[(4 - x^2)^(1/3) + Sqrt[x^2 - 3] == 1, Reals]

only returns $x=\pm2$ and $x=\pm\sqrt{3}$. Why is that?

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    $\begingroup$ Long story short: Use CubeRoot instead of ^(1/3). $\endgroup$
    – Domen
    Commented Jan 2 at 9:38

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