Suppose that x,y, and z satisfy the equations xyzx3+y3+z3xy2+x2y+xz2+x2z+yz2+y2z=4,=4,=12.Calculate the value of xy+yz+zx.
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Let s1=x+y+z and s2=xy+xz+yz. Then s1s2=(x+y+z)(xy+xz+yz)=x2y+xy2+x2z+xz2+y2z+yz2+3xyz=12+3⋅4=24.Also, s13=(x+y+z)3=(x3+y3+z3)+3(x2y+xy2+x2z+xz2+y2z+yz2)+6xyz=4+3⋅12+6⋅4=64,so s1=4. Hence, s2=s124=6.