Question

If $u=\log (x^3+y^3+z^3-3xyz)$ then $(x+y+z)(u_x+u_y+u_z)=$

• A. $0$
• B. $x-y+z$
• C. $2$
• D. $3$

Answer 1

Answer: (D)

Given:

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

Differentiate with respect to '$x$'

$\frac{du}{dx}=\frac{d}{dx}\left(\log \right(x^3+y^3+z^3-3xyz)$

[Since, $\frac{d}{dx}\left(\log x\right)=\frac{1}{x},x>0,\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ and $\frac{d}{dx}\left(\text{constant}\right)=0$]

$\Rightarrow \frac{du}{dx}=\frac{3x^2+0+0-3yz}{x^3+y^3+z^3-3xyz}$

$\Rightarrow \frac{du}{dx}=\frac{3x^2-3yz}{x^3+y^3+z^3-3xyz}$

i.e., $u_x=\frac{du}{dx}=\frac{3x^2-3yz}{x^3+y^3+z^3-3xyz}$..........(1)

Similarly,

$u_y=\frac{du}{dy}=\frac{3y^2-3xz}{x^3+y^3+z^3-3xyz}$..........(2)

$u_z=\frac{du}{dz}=\frac{3z^2-3xy}{x^3+y^3+z^3-3xyz}$..........(3)

Add all the three equations

$\frac{du}{dx}+\frac{du}{dy}+\frac{du}{dz}$

$=\frac{3x^2-3yz}{x^3+y^3+z^3-3xyz}+\frac{3y^2-3xz}{x^3+y^3+z^3-3xyz}+\frac{3z^2-3xy}{x^3+y^3+z^3-3xyz}$

$=\frac{3(x^2+y^2+z^2-xy-yz-zx)}{x^3+y^3+z^3-3xyz}$

$=\frac{3(x^2+y^2+z^2-xy-yz-zx)}{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}$

[Since, $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$]

$=\frac{3}{x+y+z}$

$\Rightarrow u_x+u_y+u_z=\frac{3}{x+y+z}$

Now, $(x+y+z)(u_x+u_y+u_z)$

$=(x+y+z)\left(\frac{3}{x+y+z}\right)$

$=3$

$\therefore (x+y+z)(u_x+u_y+u_z)=3$

Hence, Option A is correct.