Question : If $(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) = 0$ and $x+y+z = 9$, what is the value of $x^{3} + y^{3}+z^{3}-xyz$?
Option 1: 81
Option 2: 361
Option 3: 729
Option 4: 6561
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Correct Answer: 729
Solution : Given: $\left ( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right )=0$ ⇒ $xy + yz+zx = 0$ Given: $x+y+z = 9$ We know, $x^{3}+y^{3}+z^{3}-3xyz = \left ( x+y+z \right )\left ( x^{2}+y^{2}+z^{2}-xy-yz-zx \right )$ ⇒ $x^{3}+y^{3}+z^{3}-3xyz = \left ( x+y+z \right )\left (\left ( x+y+z \right )^{2}-3\left ( xy+yz+zx \right ) \right )$ Substituting values of the expressions $x+y+z$ and $xy+yz+zx$, $x^{3}+y^{3}+z^{3}-3xyz = 9\times( 9^{2}-3(0)=729$ Hence, the correct answer is 729.
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