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Question Number 50547 by Pk1167156@gmail.com last updated on 17/Dec/18

Answered by behi83417@gmail.com last updated on 17/Dec/18

35+2(xy+yz+zx)=1 ⇒xy+yz+zx=−17⇒xy=−17−z(y+x) (x+y)(x^2 +y^2 −xy)+z^3 =97 (1−z)(35−z^2 −(−17−z(1−z))=97 (1−z)(52+z−2z^2 )=97 52+z−2z^2 −52z−z^2 +2z^3 =97 ⇒2z^3 −3z^2 −51z−45=0 z=−3.72,−0.97,+6.2 ⇒1) { ((z=−3.72⇒x+y=4.72)),((xy=−17−(−3.72)(4.72)=0.56)) :} t^2 −4.72t+0.56=0⇒t=((4.72±4.48)/2) ⇒(x,y,z)=(4.6,0.12,−3.72),(0.12,4.6,−3.72) 2) { ((z=−.97⇒x+y=1.97)),((xy=−17−(−.97)(1.97)=−15.09)) :} t^2 −1.97t−15.09=0⇒t=((1.97±8.02)/2) ⇒(x,y,z)#(5,−3,−1),(−3,5,−1) 3) { ((z=6.2⇒x+y=−5.2)),((xy=−17−(6.2)(−5.2)=15.24)) :} t^2 +5.2t+15.24=0⇒t=((−5.2±5.82i)/2) ⇒(x,y,z)=(−2.6+2.91i,−2.6−2.91i,6.2), ,(−2.6−2.91i,−2.6+2.91i,6.2) .■

$$\mathrm{35}+\mathrm{2}\left({xy}+{yz}+{zx}\right)=\mathrm{1} \\ $$$$\Rightarrow{xy}+{yz}+{zx}=−\mathrm{17}\Rightarrow{xy}=−\mathrm{17}−{z}\left({y}+{x}\right) \\ $$$$\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{xy}\right)+{z}^{\mathrm{3}} =\mathrm{97} \\ $$$$\left(\mathrm{1}−{z}\right)\left(\mathrm{35}−{z}^{\mathrm{2}} −\left(−\mathrm{17}−{z}\left(\mathrm{1}−{z}\right)\right)=\mathrm{97}\right. \\ $$$$\left(\mathrm{1}−{z}\right)\left(\mathrm{52}+{z}−\mathrm{2}{z}^{\mathrm{2}} \right)=\mathrm{97} \\ $$$$\mathrm{52}+{z}−\mathrm{2}{z}^{\mathrm{2}} −\mathrm{52}{z}−{z}^{\mathrm{2}} +\mathrm{2}{z}^{\mathrm{3}} =\mathrm{97} \\ $$$$\Rightarrow\mathrm{2}{z}^{\mathrm{3}} −\mathrm{3}{z}^{\mathrm{2}} −\mathrm{51}{z}−\mathrm{45}=\mathrm{0} \\ $$$${z}=−\mathrm{3}.\mathrm{72},−\mathrm{0}.\mathrm{97},+\mathrm{6}.\mathrm{2} \\ $$$$\left.\Rightarrow\mathrm{1}\right)\begin{cases}{{z}=−\mathrm{3}.\mathrm{72}\Rightarrow{x}+{y}=\mathrm{4}.\mathrm{72}}\\{{xy}=−\mathrm{17}−\left(−\mathrm{3}.\mathrm{72}\right)\left(\mathrm{4}.\mathrm{72}\right)=\mathrm{0}.\mathrm{56}}\end{cases} \\ $$$${t}^{\mathrm{2}} −\mathrm{4}.\mathrm{72}{t}+\mathrm{0}.\mathrm{56}=\mathrm{0}\Rightarrow{t}=\frac{\mathrm{4}.\mathrm{72}\pm\mathrm{4}.\mathrm{48}}{\mathrm{2}} \\ $$$$\Rightarrow\left({x},{y},{z}\right)=\left(\mathrm{4}.\mathrm{6},\mathrm{0}.\mathrm{12},−\mathrm{3}.\mathrm{72}\right),\left(\mathrm{0}.\mathrm{12},\mathrm{4}.\mathrm{6},−\mathrm{3}.\mathrm{72}\right) \\ $$$$\left.\mathrm{2}\right)\begin{cases}{{z}=−.\mathrm{97}\Rightarrow{x}+{y}=\mathrm{1}.\mathrm{97}}\\{{xy}=−\mathrm{17}−\left(−.\mathrm{97}\right)\left(\mathrm{1}.\mathrm{97}\right)=−\mathrm{15}.\mathrm{09}}\end{cases} \\ $$$${t}^{\mathrm{2}} −\mathrm{1}.\mathrm{97}{t}−\mathrm{15}.\mathrm{09}=\mathrm{0}\Rightarrow{t}=\frac{\mathrm{1}.\mathrm{97}\pm\mathrm{8}.\mathrm{02}}{\mathrm{2}} \\ $$$$\Rightarrow\left({x},{y},{z}\right)#\left(\mathrm{5},−\mathrm{3},−\mathrm{1}\right),\left(−\mathrm{3},\mathrm{5},−\mathrm{1}\right) \\ $$$$\left.\mathrm{3}\right)\begin{cases}{{z}=\mathrm{6}.\mathrm{2}\Rightarrow{x}+{y}=−\mathrm{5}.\mathrm{2}}\\{{xy}=−\mathrm{17}−\left(\mathrm{6}.\mathrm{2}\right)\left(−\mathrm{5}.\mathrm{2}\right)=\mathrm{15}.\mathrm{24}}\end{cases} \\ $$$${t}^{\mathrm{2}} +\mathrm{5}.\mathrm{2}{t}+\mathrm{15}.\mathrm{24}=\mathrm{0}\Rightarrow{t}=\frac{−\mathrm{5}.\mathrm{2}\pm\mathrm{5}.\mathrm{82}{i}}{\mathrm{2}} \\ $$$$\Rightarrow\left({x},{y},{z}\right)=\left(−\mathrm{2}.\mathrm{6}+\mathrm{2}.\mathrm{91}{i},−\mathrm{2}.\mathrm{6}−\mathrm{2}.\mathrm{91}{i},\mathrm{6}.\mathrm{2}\right), \\ $$$$,\left(−\mathrm{2}.\mathrm{6}−\mathrm{2}.\mathrm{91}{i},−\mathrm{2}.\mathrm{6}+\mathrm{2}.\mathrm{91}{i},\mathrm{6}.\mathrm{2}\right)\:.\blacksquare \\ $$

Commented by Pk1167156@gmail.com last updated on 18/Dec/18

thank you!

Answered by mr W last updated on 17/Dec/18

⇒x+y+z=1 (x+y+z)^2 =x^2 +y^2 +z^2 +2(xy+yz+zx) 1=35+2(xy+yz+zx) ⇒xy+yz+zx=−17 (x+y+z)^3 =x^3 +y^3 +z^3 +3(xy+yz+zx)(x+y+z)−3xyz 1=97+3(−17)(1)−3xyz ⇒xyz=15 x,y,z are roots of eqn. p^3 −p^2 −17p−15=0 (p−5)(p+1)(p+3)=0 p=5, −1, −3 ⇒x,y,z ∈(5,−1,−3) ==== generally x+y+z=a x^2 +y^2 +z^2 =b x^3 +y^3 +z^3 =c ⇒xy+yz+zx=((a^2 −b)/2) ⇒xyz=((a^3 −3ab+2c)/6) x,y,z are roots of eqn. p^3 −ap^2 +((a^2 −b)/2)p−((a^3 −3ab+2c)/6)=0

$$\Rightarrow{x}+{y}+{z}=\mathrm{1} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}\left({xy}+{yz}+{zx}\right) \\ $$$$\mathrm{1}=\mathrm{35}+\mathrm{2}\left({xy}+{yz}+{zx}\right) \\ $$$$\Rightarrow{xy}+{yz}+{zx}=−\mathrm{17} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{3}} ={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} +\mathrm{3}\left({xy}+{yz}+{zx}\right)\left({x}+{y}+{z}\right)−\mathrm{3}{xyz} \\ $$$$\mathrm{1}=\mathrm{97}+\mathrm{3}\left(−\mathrm{17}\right)\left(\mathrm{1}\right)−\mathrm{3}{xyz} \\ $$$$\Rightarrow{xyz}=\mathrm{15} \\ $$$$ \\ $$$${x},{y},{z}\:{are}\:{roots}\:{of}\:{eqn}. \\ $$$${p}^{\mathrm{3}} −{p}^{\mathrm{2}} −\mathrm{17}{p}−\mathrm{15}=\mathrm{0} \\ $$$$\left({p}−\mathrm{5}\right)\left({p}+\mathrm{1}\right)\left({p}+\mathrm{3}\right)=\mathrm{0} \\ $$$${p}=\mathrm{5},\:−\mathrm{1},\:−\mathrm{3} \\ $$$$\Rightarrow{x},{y},{z}\:\in\left(\mathrm{5},−\mathrm{1},−\mathrm{3}\right) \\ $$$$ \\ $$$$==== \\ $$$${generally} \\ $$$${x}+{y}+{z}={a} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={b} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} ={c} \\ $$$$\Rightarrow{xy}+{yz}+{zx}=\frac{{a}^{\mathrm{2}} −{b}}{\mathrm{2}} \\ $$$$\Rightarrow{xyz}=\frac{{a}^{\mathrm{3}} −\mathrm{3}{ab}+\mathrm{2}{c}}{\mathrm{6}} \\ $$$${x},{y},{z}\:{are}\:{roots}\:{of}\:{eqn}. \\ $$$${p}^{\mathrm{3}} −{ap}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} −{b}}{\mathrm{2}}{p}−\frac{{a}^{\mathrm{3}} −\mathrm{3}{ab}+\mathrm{2}{c}}{\mathrm{6}}=\mathrm{0} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18

excellent....

$${excellent}.... \\ $$

Commented by afachri last updated on 17/Dec/18

Amazing. Inpired thinker you are, Sir.

$$\mathrm{Amazing}.\:\mathrm{Inpired}\:\mathrm{thinker}\:\mathrm{you}\:\mathrm{are},\:\mathrm{Sir}. \\ $$

Commented by OTCHRRE ABDULLAI last updated on 17/Dec/18

My great man that

$${My}\:{great}\:{man}\:{that} \\ $$

Commented by Necxx last updated on 17/Dec/18

i posted this same question 2 years and mrW solved it. To my greatest surprise the style he used is totally different from now. MrW is indeed a deep thinker.

$${i}\:{posted}\:{this}\:{same}\:{question}\:\mathrm{2}\:{years} \\ $$$${and}\:{mrW}\:{solved}\:{it}.\:{To}\:{my}\:{greatest} \\ $$$${surprise}\:{the}\:{style}\:{he}\:{used}\:{is}\: \\ $$$${totally}\:{different}\:{from}\:{now}. \\ $$$${MrW}\:{is}\:{indeed}\:{a}\:{deep}\:{thinker}. \\ $$

Commented by Pk1167156@gmail.com last updated on 18/Dec/18

great sir!

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