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x-y-z-1-i-x-2-y-2-z-2-37-ii-x-3-y-2-z-3-91-iii-Solve-simultaneously-

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Question Number 7809 by Tawakalitu. last updated on 16/Sep/16
x + y + z = 1 ......... (i) x^2 + y^2 + z^2 = 37 ........ (ii) x^3 + y^2 + z^3 = 91 ........ (iii) Solve simultaneously.
$${x}\:+\:{y}\:+\:{z}\:=\:\mathrm{1}\:\:\:\:\:\:………\:\left({i}\right) \\ $$$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \:=\:\mathrm{37}\:\:\:\:\:……..\:\left({ii}\right) \\ $$$${x}^{\mathrm{3}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{3}} \:=\:\mathrm{91}\:\:\:\:\:……..\:\left({iii}\right) \\ $$$$ \\ $$$${Solve}\:{simultaneously}.\: \\ $$
Commented by Tawakalitu. last updated on 16/Sep/16
That is the correct question , no mistake in the last equation ....thanks for your help.
$${That}\:{is}\:{the}\:{correct}\:{question}\:,\:{no}\:{mistake}\:{in}\:{the}\:{last} \\ $$$${equation}\:….{thanks}\:{for}\:{your}\:{help}. \\ $$
Commented by prakash jain last updated on 16/Sep/16
x^3 +z^3 +y^2 =91 (x+z)(x^2 +z^2 −xz)+y^2 =91 from (i) x+z=1−y from (ii) x^2 +z^2 =37−y^2 (1−y)(37−y^2 −xz)+y^2 =91 (iv) from (ii) x^2 +z^2 +2xz−2xz+y^2 =37 (x+z)^2 −2xz+y^2 =37 (1−y)^2 −2xz+y^2 =37⇒xz=(((1−y)^2 +y^2 −37)/2) from (iv) (1−y)(37−y^2 −(((1−y)^2 +y^2 −37)/2))+y^2 =91 (1−y)(((74−2y^2 −1−y^2 +2y−y^2 +37)/2))+y^2 =91 (1−y)(110+2y)+2y^2 =182 110−110y+2y−2y^2 +2y^2 =182 110−108y=182 55−54y=91 −54y=91−55=36⇒y=−((18)/(27))=−(2/3)
$${x}^{\mathrm{3}} +{z}^{\mathrm{3}} +{y}^{\mathrm{2}} =\mathrm{91} \\ $$$$\left({x}+{z}\right)\left({x}^{\mathrm{2}} +{z}^{\mathrm{2}} −{xz}\right)+{y}^{\mathrm{2}} =\mathrm{91} \\ $$$${from}\:\left({i}\right)\:{x}+{z}=\mathrm{1}−{y} \\ $$$${from}\:\left({ii}\right)\:{x}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{37}−{y}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}−{y}\right)\left(\mathrm{37}−{y}^{\mathrm{2}} −{xz}\right)+{y}^{\mathrm{2}} =\mathrm{91}\:\left({iv}\right) \\ $$$${from}\:\left({ii}\right) \\ $$$${x}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}{xz}−\mathrm{2}{xz}+{y}^{\mathrm{2}} =\mathrm{37} \\ $$$$\left({x}+{z}\right)^{\mathrm{2}} −\mathrm{2}{xz}+{y}^{\mathrm{2}} =\mathrm{37} \\ $$$$\left(\mathrm{1}−{y}\right)^{\mathrm{2}} −\mathrm{2}{xz}+{y}^{\mathrm{2}} =\mathrm{37}\Rightarrow{xz}=\frac{\left(\mathrm{1}−{y}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{37}}{\mathrm{2}} \\ $$$${from}\:\left({iv}\right) \\ $$$$\left(\mathrm{1}−{y}\right)\left(\mathrm{37}−{y}^{\mathrm{2}} −\frac{\left(\mathrm{1}−{y}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{37}}{\mathrm{2}}\right)+{y}^{\mathrm{2}} =\mathrm{91} \\ $$$$\left(\mathrm{1}−{y}\right)\left(\frac{\mathrm{74}−\mathrm{2}{y}^{\mathrm{2}} −\mathrm{1}−{y}^{\mathrm{2}} +\mathrm{2}{y}−{y}^{\mathrm{2}} +\mathrm{37}}{\mathrm{2}}\right)+{y}^{\mathrm{2}} =\mathrm{91} \\ $$$$\left(\mathrm{1}−{y}\right)\left(\mathrm{110}+\mathrm{2}{y}\right)+\mathrm{2}{y}^{\mathrm{2}} =\mathrm{182} \\ $$$$\mathrm{110}−\mathrm{110}{y}+\mathrm{2}{y}−\mathrm{2}{y}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} =\mathrm{182} \\ $$$$\mathrm{110}−\mathrm{108}{y}=\mathrm{182} \\ $$$$\mathrm{55}−\mathrm{54}{y}=\mathrm{91} \\ $$$$−\mathrm{54}{y}=\mathrm{91}−\mathrm{55}=\mathrm{36}\Rightarrow{y}=−\frac{\mathrm{18}}{\mathrm{27}}=−\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Commented by prakash jain last updated on 16/Sep/16
xz=y^2 −y−18=(4/9)+(2/3)−18 x+z=1−y=(5/3) x+z and xz are known so x−z can be found and value of x and z can be obtained as well.
$${xz}={y}^{\mathrm{2}} −{y}−\mathrm{18}=\frac{\mathrm{4}}{\mathrm{9}}+\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{18} \\ $$$${x}+{z}=\mathrm{1}−{y}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$${x}+{z}\:\mathrm{and}\:{xz}\:{are}\:{known}\:{so}\:{x}−{z}\:{can}\:{be}\:{found} \\ $$$${and}\:{value}\:{of}\:{x}\:{and}\:{z}\:{can}\:{be}\:{obtained}\:{as}\:{well}. \\ $$
Answered by Rasheed Soomro last updated on 18/Sep/16
x + y + z = 1 ......... (i) x^2 + y^2 + z^2 = 37 ........ (ii) x^3 + y^2 + z^3 = 91 ........ (iii) (i)⇒x+z=1−y (ii)⇒x^2 +z^2 =37−y^2 ⇒(x+z)^2 −2xz=37−y^2 ⇒(1−y)^2 −2xz=37−y^2 ⇒(1−y)^2 +y^2 −37=2xz ⇒xz=((1−2y+y^2 +y^2 −37)/2)=((2y^2 −2y−36)/2)=y^2 −y−18...(iv) (iii)⇒x^3 +z^3 =91−y^2 ⇒(x+z)^3 −3xz(x+z)=91−y^2 ⇒(1−y)^3 −3xz(1−y)=91−y^2 ⇒xz=(((1−y)^3 +y^2 −91)/(3(1−y)))..................................(v) (iv),(v)⇒y^2 −y−18=(((1−y)^3 +y^2 −91)/(3(1−y))) See the comments(By prakash jain) for complete answer.
$${x}\:+\:{y}\:+\:{z}\:=\:\mathrm{1}\:\:\:\:\:\:………\:\left({i}\right) \\ $$$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \:=\:\mathrm{37}\:\:\:\:\:……..\:\left({ii}\right) \\ $$$${x}^{\mathrm{3}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{3}} \:=\:\mathrm{91}\:\:\:\:\:……..\:\left({iii}\right) \\ $$$$\left({i}\right)\Rightarrow{x}+{z}=\mathrm{1}−{y} \\ $$$$\left({ii}\right)\Rightarrow{x}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{37}−{y}^{\mathrm{2}} \Rightarrow\left({x}+{z}\right)^{\mathrm{2}} −\mathrm{2}{xz}=\mathrm{37}−{y}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\left(\mathrm{1}−{y}\right)^{\mathrm{2}} −\mathrm{2}{xz}=\mathrm{37}−{y}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\left(\mathrm{1}−{y}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{37}=\mathrm{2}{xz} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow{xz}=\frac{\mathrm{1}−\mathrm{2}{y}+{y}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{37}}{\mathrm{2}}=\frac{\mathrm{2}{y}^{\mathrm{2}} −\mathrm{2}{y}−\mathrm{36}}{\mathrm{2}}={y}^{\mathrm{2}} −{y}−\mathrm{18}…\left({iv}\right) \\ $$$$\left({iii}\right)\Rightarrow{x}^{\mathrm{3}} +{z}^{\mathrm{3}} =\mathrm{91}−{y}^{\mathrm{2}} \Rightarrow\left({x}+{z}\right)^{\mathrm{3}} −\mathrm{3}{xz}\left({x}+{z}\right)=\mathrm{91}−{y}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\left(\mathrm{1}−{y}\right)^{\mathrm{3}} −\mathrm{3}{xz}\left(\mathrm{1}−{y}\right)=\mathrm{91}−{y}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow{xz}=\frac{\left(\mathrm{1}−{y}\right)^{\mathrm{3}} +{y}^{\mathrm{2}} −\mathrm{91}}{\mathrm{3}\left(\mathrm{1}−{y}\right)}…………………………….\left({v}\right) \\ $$$$\left({iv}\right),\left({v}\right)\Rightarrow{y}^{\mathrm{2}} −{y}−\mathrm{18}=\frac{\left(\mathrm{1}−{y}\right)^{\mathrm{3}} +{y}^{\mathrm{2}} −\mathrm{91}}{\mathrm{3}\left(\mathrm{1}−{y}\right)} \\ $$$${See}\:{the}\:{comments}\left({By}\:{prakash}\:{jain}\right)\:{for}\:{complete}\:{answer}. \\ $$

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