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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CubeRootinstead of^(1/3). – Domen Jan 02 '24 at 09:38