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Question Number 109307 by john santu last updated on 22/Aug/20

Answered by mr W last updated on 22/Aug/20

x^2 +2yz−2(y+z)=82 y^2 +2zx−2(z+x)=83 z^2 +2xy−2(x+y)=87 Σ: (x+y+z)^2 −4(x+y+z)=252 (x+y+z)^2 −4(x+y+z)+4=256 (x+y+z−2)^2 =16^2 x+y+z−2=±16 ⇒x+y+z=2±16=18 or −14

x2+2yz−2(y+z)=82y2+2zx−2(z+x)=83z2+2xy−2(x+y)=87Σ:(x+y+z)2−4(x+y+z)=252(x+y+z)2−4(x+y+z)+4=256(x+y+z−2)2=162x+y+z−2=±16⇒x+y+z=2±16=18or−14

Commented by bobhans last updated on 23/Aug/20

((awesome−−−−nice−−−)/)

awesome−−−−nice−−−

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