x-y-z-gt-0-Solve-for-real-numbers-x-3-y-3-z-3-3-x-1-3-y-1-3-z-1-3-12-x-y-z-1-

Question Number 158363 by HongKing last updated on 03/Nov/21

x;y;z>0 Solve for real numbers: { ((x^3 + y^3 + z^3 + 3∙((x)^(1/3) + (y)^(1/3) + (z)^(1/3) ) = 12)),((x∙y∙z = 1)) :}

$$\mathrm{x};\mathrm{y};\mathrm{z}>\mathrm{0} \\ $$$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\begin{cases}{\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{z}^{\mathrm{3}} \:+\:\mathrm{3}\centerdot\left(\sqrt[{\mathrm{3}}]{\mathrm{x}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{y}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{z}}\right)\:=\:\mathrm{12}}\\{\mathrm{x}\centerdot\mathrm{y}\centerdot\mathrm{z}\:=\:\mathrm{1}}\end{cases} \\ $$$$ \\ $$

Answered by GuruBelakangPadang last updated on 03/Nov/21

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

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