Question Number 158363 by HongKing last updated on 03/Nov/21
$$\mathrm{x};\mathrm{y};\mathrm{z}>\mathrm{0} \\ $$$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\begin{cases}{\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{z}^{\mathrm{3}} \:+\:\mathrm{3}\centerdot\left(\sqrt[{\mathrm{3}}]{\mathrm{x}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{y}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{z}}\right)\:=\:\mathrm{12}}\\{\mathrm{x}\centerdot\mathrm{y}\centerdot\mathrm{z}\:=\:\mathrm{1}}\end{cases} \\ $$$$ \\ $$
Answered by GuruBelakangPadang last updated on 03/Nov/21
$${simetrice}:{x}={y}={z}=\mathrm{1} \\ $$