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

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

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$3$

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(D) Given, $u=\log \left(x^3+y^3+z^3-3 x y z\right)$.We know that $x^3+y^3+z^3-3 x y z = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$.So, $u = \log(x+y+z) + \log(x^2+y^2+z^2-xy-yz-zx)$.Now, calculate the partial derivatives:$u_x = \frac{\partial u}{\partial x} = \frac{3x^2-3yz}{x^3+y^3+z^3-3xyz}$,$u_y = \frac{\partial u}{\partial y} = \frac{3y^2-3xz}{x^3+y^3+z^3-3xyz}$,$u_z = \frac{\partial u}{\partial z} = \frac{3z^2-3xy}{x^3+y^3+z^3-3xyz}$.Summing these up:$u_x+u_y+u_z = \frac{3(x^2+y^2+z^2-xy-yz-zx)}{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)} = \frac{3}{x+y+z}$.Therefore, $(x+y+z)(u_x+u_y+u_z) = 3$.

If $u=\log \left(x^3+y^3+z^3-3 x y z\right)$, then $(x+y+z)(u_x+u_y+u_z)$ is equal to

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