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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=1………(i)x2+y2+z2=37……..(ii)x3+y2+z3=91……..(iii)Solvesimultaneously.
Commented by Tawakalitu. last updated on 16/Sep/16
That is the correct question , no mistake in the last equation ....thanks for your help.
Thatisthecorrectquestion,nomistakeinthelastequation….thanksforyourhelp.
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)
x3+z3+y2=91(x+z)(x2+z2−xz)+y2=91from(i)x+z=1−yfrom(ii)x2+z2=37−y2(1−y)(37−y2−xz)+y2=91(iv)from(ii)x2+z2+2xz−2xz+y2=37(x+z)2−2xz+y2=37(1−y)2−2xz+y2=37⇒xz=(1−y)2+y2−372from(iv)(1−y)(37−y2−(1−y)2+y2−372)+y2=91(1−y)(74−2y2−1−y2+2y−y2+372)+y2=91(1−y)(110+2y)+2y2=182110−110y+2y−2y2+2y2=182110−108y=18255−54y=91−54y=91−55=36⇒y=−1827=−23
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=y2−y−18=49+23−18x+z=1−y=53x+zandxzareknownsox−zcanbefoundandvalueofxandzcanbeobtainedaswell.
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=1………(i)x2+y2+z2=37……..(ii)x3+y2+z3=91……..(iii)(i)⇒x+z=1−y(ii)⇒x2+z2=37−y2⇒(x+z)2−2xz=37−y2⇒(1−y)2−2xz=37−y2⇒(1−y)2+y2−37=2xz⇒xz=1−2y+y2+y2−372=2y2−2y−362=y2−y−18…(iv)(iii)⇒x3+z3=91−y2⇒(x+z)3−3xz(x+z)=91−y2⇒(1−y)3−3xz(1−y)=91−y2⇒xz=(1−y)3+y2−913(1−y)…………………………….(v)(iv),(v)⇒y2−y−18=(1−y)3+y2−913(1−y)Seethecomments(Byprakashjain)forcompleteanswer.

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