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    Question

    x, y and z are real numbers. If x3Β +

    y3Β + z3Β = 13,x + y + z = 1 and xyz = 1, then what is the value of xy + yz + zx?
    A -1 Correct Answer Incorrect Answer
    B 1 Correct Answer Incorrect Answer
    C 3 Correct Answer Incorrect Answer
    D -3 Correct Answer Incorrect Answer

    Solution

    x3 + y3 + z3βˆ’ 3xyz (x + y + z)(x2 + y2 + z2 βˆ’ xy βˆ’ yz βˆ’ zx) x3 + y3 + z3 βˆ’ 3xyz (x + y + z)((x + y + z)2 βˆ’ 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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