If u = ( x^3 + y^3 + z^3 - 3xyz), then ( u x + u y + u z ) (x + y… - Vidyadip

Question

If $u = \log ({x^3} + {y^3} + {z^3} - 3xyz)$, then

$\left( {{{\partial u} \over {\partial x}} + {{\partial u} \over {\partial y}} + {{\partial u} \over {\partial z}}} \right)$ $(x + y + z) =$

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Answer

d
(d) $u = \log ({x^3} + {y^3} + {z^3} - 3xyz)$

$\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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