If u = log (x^3 + y^3 + z^3 - 3xyz),then left | vedclass

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(D) Given $u = \log (x^3 + y^3 + z^3 - 3xyz)$.We know that $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$.Therefore,$u = \log(

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$3$

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(D) Given $u = \log (x^3 + y^3 + z^3 - 3xyz)$.We know that $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$.Therefore,$u = \log(x + y + z) + \log(x^2 + y^2 + z^2 - xy - yz - zx)$.Calculating the partial derivatives:$\frac{\partial u}{\partial x} = \frac{3x^2 - 3yz}{x^3 + y^3 + z^3 - 3xyz}$$\frac{\partial u}{\partial y} = \frac{3y^2 - 3zx}{x^3 + y^3 + z^3 - 3xyz}$$\frac{\partial u}{\partial z} = \frac{3z^2 - 3xy}{x^3 + y^3 + z^3 - 3xyz}$Summing these derivatives:$\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}$Multiplying by $(x + y + z)$:$(x + y + z) \left( \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} \right) = 3$.

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

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