Question

If $u=\log \left(x^{3}+y^{3}+z^{3}-3 x y z\right)$, then $(x +y +z)\left(u_{x}+u_{y}+u_{z}\right)$ is equal to

• A. 0
• B. x - y + z
• C. 2
• D. 3

Answer 1

Answer: (D)

Given, $u=\log \left(x^{3}+y^{3}+z^{3}-3 x y z\right)$
$u_{x}=\frac{d u}{d x}=\frac{3 x^{2}-3 y z}{\left(x^{3}+y^{3}+z^{3}-3 x y z\right)}$
$u_{y}=\frac{d u}{d y}=\frac{3 y^{2}-3 x z}{x^{3}+y^{3}+z^{3}-3 x y z}$
and $u_{z}=\frac{d u}{d z}=\frac{3 z^{2}-3 x y}{x^{3}+y^{3}+z^{3}-3 x y z}$
$u_{x}+u_{y}+u_{z}$
$=\frac{3\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)}{(x + y +z)\left(x^{2}+ y^{2}+z^{2}-x y-y z-z x\right)}$
$\Rightarrow (x +y +z)\left(u_{x}+u_{y}+u_{z}\right)=3$