MCQ
If $u = \log ({x^3} + {y^3} + {z^3} - 3xyz)$, then
$\left( {{{\partial u} \over {\partial x}} + {{\partial u} \over {\partial y}} + {{\partial u} \over {\partial z}}} \right)$ $(x + y + z) =$
- A$0$
- B$1$
- C$2$
- ✓$3$
$\left( {{{\partial u} \over {\partial x}} + {{\partial u} \over {\partial y}} + {{\partial u} \over {\partial z}}} \right)$ $(x + y + z) =$
$\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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