Question Number 153093 by naka3546 last updated on 04/Sep/21
$${Find}\:\:{the}\:\:{solution}\:\:{of}\:\:{three}\:\:{variables}\:\:{equality}\:\:{system}\:\:{x},\:{y},\:{z}\:. \\ $$$$\:\:{a}^{\mathrm{3}} \:+\:{a}^{\mathrm{2}} {x}\:+\:{ay}\:+\:{z}\:=\:\mathrm{0} \\ $$$$\:\:{b}^{\mathrm{3}} \:+\:{b}^{\mathrm{2}} {x}\:+\:{by}\:+\:{z}\:=\:\mathrm{0} \\ $$$$\:\:{c}^{\mathrm{3}} \:+\:{c}^{\mathrm{2}} {x}\:+\:{cy}\:+\:{z}\:=\:\mathrm{0} \\ $$$$ \\ $$$${Thank}\:\:{you}\:\:{so}\:\:{much} \\ $$
Commented by MJS_new last updated on 04/Sep/21
$$\mathrm{it}'\mathrm{s}\:\mathrm{linear}\:\mathrm{for}\:{x},\:{y},\:{z} \\ $$$${x}=−\left({a}+{b}+{c}\right) \\ $$$${y}={ab}+{ac}+{bc} \\ $$$${z}=−{abc} \\ $$
Answered by mr W last updated on 05/Sep/21
$${a},{b},{c}\:{are}\:{the}\:{roots}\:{of}\:\:{following}\: \\ $$$${cubic}\:{eqn}.\:{w}.{r}.{t}.\:{t}: \\ $$$${t}^{\mathrm{3}} +{xt}^{\mathrm{2}} +{yt}+{z}=\mathrm{0} \\ $$$$\Rightarrow{a}+{b}+{c}=−{x}\:\Rightarrow{x}=−\left({a}+{b}+{c}\right) \\ $$$$\Rightarrow{ab}+{bc}+{ca}={y}\:\Rightarrow{y}={ab}+{bc}+{ca} \\ $$$$\Rightarrow{abc}=−{z}\:\Rightarrow{z}=−{abc} \\ $$