A) 0
B) 1
C) 2
D) 3
Correct Answer: D
Solution :
\[u=\log ({{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz)\] \[\therefore \]\[\frac{\partial u}{\partial x}=\frac{3{{x}^{2}}-3yz}{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz}\];\[\frac{\partial u}{\partial y}=\frac{3{{y}^{2}}-3zx}{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz}\] \[\frac{\partial u}{\partial z}=\frac{3{{z}^{2}}-3xy}{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz}\] \[\therefore \]\[\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}\]=\[\frac{3\,({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)}{(x+y+z)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)}\] =\[\frac{3}{x+y+z}\,\]. \[\therefore \]\[\,(x+y+z)\,\left( \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z} \right)=3\].
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