Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is D

$3$

Explanation for the correct answer:

Find the value of $\begin{array}{rcl}& & \left(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}\right)\end{array}(x+y+z)$ :

Given,

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

Now, differentiate partially with respect to $x,y$and $z$

$$\begin{gathered} \frac{\partial u}{\partial x}=\frac{\left(3x^2–3yz\right)}{(x^3+y^3+z^3–3xyz)}...\left(i\right) \\ \frac{\partial u}{\partial y}=\frac{(3y^2–3xz)}{(x^3+y^3+z^3–3xyz)}...\left(ii\right) \\ \frac{\partial u}{\partial z}=\frac{(3z^2–3xy)}{(x^3+y^3+z^3–3xyz)}...\left(iii\right) \end{gathered}$$

Adding the above three equations,

$\begin{array}{rcl}\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}& =& \frac{(3x^2+3y^2+3z^2–3yz–3xz–3xy)}{(x^3+y^3+z^3–3xyz)}\\ & =& \frac{3(x^2+y^2+z^2–yz–xz–xy)}{(x^3+y^3+z^3–3xyz)}\end{array}$

We know,

$(x^3+y^3+z^3–3xyz)=(x+y+z)(x^2+y^2+z^2–yz–xz–xy)$

Then,

$\begin{array}{rcl}\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}& =& \frac{3(x^2+y^2+z^2–yz–xz–xy)}{(x+y+z)(x^2+y^2+z^2–yz–xz–xy)}\\ & =& \frac{3}{(x+y+z)}\\ & \Rightarrow & \left(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}\right)(x+y+z)=3\end{array}$

Hence, the correct option is D.