A) 0
B) 1
C) \[u\]
D) 3
E) \[-1\]
Correct Answer: D
Solution :
\[\because \]\[u=\log ({{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz)\] On differentiating w. r. t.\[x,y,z\]respectively, we get \[\frac{\partial u}{\partial x}=\frac{1}{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz}(3{{x}^{2}}-3yz)\] \[\frac{\partial u}{\partial y}=\frac{1}{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz}(3{{y}^{2}}-3xz)\] and \[\frac{\partial u}{\partial z}=\frac{1}{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz}(3{{z}^{2}}-3xy)\] \[\therefore \] \[(x+y+z)\left( \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z} \right)\] \[=\frac{3(x+y+z)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)}{({{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz)}\] \[=\frac{3({{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz)}{({{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz)}=3\]
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