Question

If $u=\log (x^3+y^3+z^3-3xyz)$ and ${\left(\frac{\partial }{\partial x}+\frac{\partial }{\partial y}+\frac{\partial }{\partial z}\right)}^{2}u=\frac{-k}{(x+y+z{)}^2}$

then $k=?$

• A. $6$
• B. $3$
• C. $9$
• D. $5$

Answer 1

Answer: (C)

Given:

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

$\Rightarrow \frac{\partial u}{\partial x}=\frac{\partial \left(\log \right(x^3+y^3+z^3-3xyz\left)\right)}{\partial x}$

Here, recall that,

$\frac{d}{dx}\left(\log x\right)=\frac{1}{x},x>0$

$\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

$\frac{d}{dx}\left(\text{constant}\right)=0$

$\Rightarrow \frac{\partial u}{\partial x}=\frac{1}{\log (x^3+y^3+z^3-3xyz)}(3x^2+0+0-3yz)$

$\Rightarrow \frac{\partial u}{\partial x}=\frac{3x^2-3yz}{\log (x^3+y^3+z^3-3xyz)}$ ---(1)

Similarly,

$\Rightarrow \frac{\partial u}{\partial y}=\frac{3y^2-3xz}{\log (x^3+y^3+z^3-3xyz)}$---(2)

$\Rightarrow \frac{\partial u}{\partial z}=\frac{3z^2-3xy}{\log (x^3+y^3+z^3-3xyz)}$---(3)

Now,

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

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

[$\because x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy+yz+xz)$]
$\Rightarrow \frac{\partial u}{dx}+\frac{\partial u}{dy}+\frac{\partial u}{dz}=\frac{3}{(x+y+z)}$

$\Rightarrow (\frac{\partial }{dx}+\frac{\partial }{dy}+\frac{\partial }{dz})u=\frac{3}{(x+y+z)}$
$\Rightarrow (\frac{\partial }{dx}+\frac{\partial }{dy}+\frac{\partial }{dz})(\frac{\partial u}{dx}+\frac{\partial u}{dy}+\frac{\partial u}{dz})={(\frac{\partial }{\partial x}+\frac{\partial }{\partial y}+\frac{\partial }{\partial z})}^{2}u$
$\Rightarrow \frac{-3}{(x+y+z{)}^2}+\frac{-3}{(x+y+z{)}^2}+\frac{-3}{(x+y+z{)}^2}={(\frac{\partial }{\partial x}+\frac{\partial }{\partial y}+\frac{\partial }{\partial z})}^{2}u$
$\Rightarrow \frac{-9}{(x+y+z{)}^2}={(\frac{\partial }{\partial x}+\frac{\partial }{\partial y}+\frac{\partial }{\partial z})}^{2}u$ ---(4)

It is given that,

$\frac{-k}{(x+y+z{)}^2}={(\frac{\partial }{\partial x}+\frac{\partial }{\partial y}+\frac{\partial }{\partial z})}^{2}u$

$\Rightarrow k=9$

Hence, Option C is correct.