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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