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Question 16
$$x, y$$ and $$z$$ are real numbers. If $$x^3 + y^3 + z^3 = 13,x + y + z = 1$$ and $$xyz = 1$$, then what is the value of $$xy + yz + zx?$$
Solution
$$x^3 + y^3 + z^3 -3xyz$$=$$(x + y + z )(x^{2}+y^{2}+z^{2}-xy-yz-zx)$$
$$x^3 + y^3 + z^3 -3xyz$$=$$(x + y + z )((x+y+z)^{2}-3(xy+yz+zx)$$
13-3(1)=(1)(1-3((xy+yz+zx)))
10=1-3(xy+yz+zx)
(xy+yz+zx)=-3
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