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 | ||
$${simetrice}:{x}={y}={z}=\mathrm{1} \\ $$ | ||