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Question Number 58789 by ANTARES VY last updated on 30/Apr/19

Commented by ANTARES VY last updated on 30/Apr/19

E=?

E=?

Commented by tanmay last updated on 30/Apr/19

E=(1/(x+yz))+(1/(y+xz))+(1/(z+xy)) =(1/(x(x+y+z)+yz))+(1/(y(x+y+z)+xz))+(1/(z(x+y+z)+xy)) =(1/(x^2 +xy+xz+yz))+(1/(xy+y^2 +yz+xz))+(1/(xz+yz+z^2 +xy)) =(1/(x(x+y)+z(x+y)))+(1/(y(x+y)+z(x+y)))+(1/(z(x+z)+y(x+z))) =(1/((x+y)(x+z)))+(1/((x+y)(y+z)))+(1/((x+z)(z+y))) =((y+z+z+x+x+y)/((x+y)(y+z)(z+x))) =((2(x+y+z))/((x+y)(y+z)(z+x))) (x+y+z)^3 =x^3 +y^3 +z^3 +3(x+y)(y+z)(z+x) 1−4=3(x+y)(y+z)(z+x) (x+y)(y+z)(z+x)=−1 hence answer is =((2(x+y+z))/((x+y)(y+z)(z+x))) =((2×1)/(−1))=−2

E=1x+yz+1y+xz+1z+xy=1x(x+y+z)+yz+1y(x+y+z)+xz+1z(x+y+z)+xy=1x2+xy+xz+yz+1xy+y2+yz+xz+1xz+yz+z2+xy=1x(x+y)+z(x+y)+1y(x+y)+z(x+y)+1z(x+z)+y(x+z)=1(x+y)(x+z)+1(x+y)(y+z)+1(x+z)(z+y)=y+z+z+x+x+y(x+y)(y+z)(z+x)=2(x+y+z)(x+y)(y+z)(z+x)(x+y+z)3=x3+y3+z3+3(x+y)(y+z)(z+x)1−4=3(x+y)(y+z)(z+x)(x+y)(y+z)(z+x)=−1henceansweris=2(x+y+z)(x+y)(y+z)(z+x)=2×1−1=−2

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