x, y and z are real numbers. If x3 + y3 + z3 = 13,x + y + z = 1 and xyz = 1, then what is the value of xy + yz + zx?
x3 + y3 + z3− 3xyz (x + y + z)(x2 + y2 + z2 − xy − yz − zx) x3 + y3 + z3 − 3xyz (x + y + z)((x + y + z)2 − 3(xy + yz + zx) 13-3(1)=(1)(1-3((xy+yz+zx))) 10=1-3(xy+yz+zx) (xy+yz+zx)= – 3
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