Question

If $u=\log \left(x^3+y^3+z^3-3xyz\right)$, 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, $u=\log \left(x^3+y^3+z^3-3xyz\right)$
$u_x=\frac{du}{dx}=\frac{3x^2-3yz}{\left(x^3+y^3+z^3-3xyz\right)}$
$u_y=\frac{du}{dy}=\frac{3y^2-3xz}{x^3+y^3+z^3-3xyz}$
and $u_z=\frac{du}{dz}=\frac{3z^2-3xy}{x^3+y^3+z^3-3xyz}$
$u_x+u_y+u_z$
$=\frac{3\left(x^2+y^2+z^2-xy-yz-zx\right)}{(x+y+z)\left(x^2+y^2+z^2-xy-yz-zx\right)}$
$\Rightarrow (x+y+z)\left(u_x+u_y+u_z\right)=3$