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Question Number 35152 by Victor31926 last updated on 16/May/18

Commented by Rasheed.Sindhi last updated on 16/May/18

x+y+z=1...............I x^2 +y^2 +z^2 =35..........II x^3 +y^3 +z^3 =97..........III II⇒(x+y+z)^2 −2xy−2yz−2zx=35 ⇒1−2xy−2yz−2zx=35 ⇒xy+yz+zx=−17 III⇒x^3 +y^3 +z^3 −3xyz=97−3xyz (x+y+z)(x^2 +y^2 +z^2 −xy−yz−zx) =97−3xyz (1)(35−(−17))=97−3xyz xyz=(97−52)/3=15.........IV I: x+y+z=1 IV: xyz=15 Assuming x,y,z∈Z xyz=15 1.3.5=15 but 1+3+5≠1 Some numbers must be −ve and in order to make the product +ve we must take any two of x,y,z negative. So we have (−1)(−3)(5)=15 (−1)+(−3)+(5)=1 Hence {x,y,z}={−1,−3,5}

x+y+z=1...............Ix2+y2+z2=35..........IIx3+y3+z3=97..........IIIII⇒(x+y+z)2−2xy−2yz−2zx=35⇒1−2xy−2yz−2zx=35⇒xy+yz+zx=−17III⇒x3+y3+z3−3xyz=97−3xyz(x+y+z)(x2+y2+z2−xy−yz−zx)=97−3xyz(1)(35−(−17))=97−3xyzxyz=(97−52)/3=15.........IVI:x+y+z=1IV:xyz=15Assumingx,y,z∈Zxyz=151.3.5=15but1+3+5≠1Somenumbersmustbe−veandinordertomaketheproduct+vewemusttakeanytwoofx,y,znegative.Sowehave(−1)(−3)(5)=15(−1)+(−3)+(5)=1Hence{x,y,z}={−1,−3,5}

Commented by behi83417@gmail.com last updated on 16/May/18

xy+z(x+y)=−17⇒((15)/z)+z(1−z)=−17 ⇒z^3 −z^2 −17z−15=0 (z+1)(z^2 −2z−15)=0⇒z=−1,−3,5 (x,y,z)=(−1,−3,5),(−3,5,−1),(5,−1,−3).

xy+z(x+y)=−17⇒15z+z(1−z)=−17⇒z3−z2−17z−15=0(z+1)(z2−2z−15)=0⇒z=−1,−3,5(x,y,z)=(−1,−3,5),(−3,5,−1),(5,−1,−3).

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