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Find-the-solution-of-three-variables-equality-system-x-y-z-a-3-a-2-x-ay-z-0-b-3-b-2-x-by-z-0-c-3-c-2-x-cy-z-0-Thank-you-so-much-




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

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