Question

$\rm If \: u=log(x^3 + {y}^{3} + {z}^{3} - 3xyz) \\ \rm then \: show \: that : - \\ \rm \bigg( \frac{ \partial}{ \partial x} + \frac{ \partial}{ \partial y} + \frac{ \partial}{ \partial z} \bigg) ^2u = \frac{ - 9}{(x + y + z) {}^{2} }$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: u=log(x^3 + {y}^{3} + {z}^{3} - 3xyz)$

On differentiating partially w. r. t., we get

$\rm \: \dfrac{\partial }{\partial x}u=\dfrac{\partial }{\partial x}log(x^3 + {y}^{3} + {z}^{3} - 3xyz)$

$\rm \: \dfrac{\partial u}{\partial x}=\dfrac{1}{x^3 + {y}^{3} + {z}^{3} - 3xyz} \dfrac{\partial }{\partial x}(x^3 + {y}^{3} + {z}^{3} - 3xyz)$

$\rm\implies \:\rm \: \dfrac{\partial u}{\partial x}=\dfrac{ {3x}^{2} - 3yz}{x^3 + {y}^{3} + {z}^{3} - 3xyz} - - - (1)$

Again,

$\rm \: u=log(x^3 + {y}^{3} + {z}^{3} - 3xyz)$

On differentiating partially w. r. t. y, we get

$\rm \: \dfrac{\partial }{\partial y}u=\dfrac{\partial }{\partial y}log(x^3 + {y}^{3} + {z}^{3} - 3xyz)$

$\rm \: \dfrac{\partial u}{\partial y}=\dfrac{1}{x^3 + {y}^{3} + {z}^{3} - 3xyz} \dfrac{\partial }{\partial y}(x^3 + {y}^{3} + {z}^{3} - 3xyz)$

$\rm\implies \:\rm \: \dfrac{\partial u}{\partial y}=\dfrac{ {3y}^{2} - 3xz}{x^3 + {y}^{3} + {z}^{3} - 3xyz} - - - (2)$

Again,

$\rm \: u=log(x^3 + {y}^{3} + {z}^{3} - 3xyz)$

On differentiating partially w. r. t. z, we get

$\rm \: \dfrac{\partial u}{\partial z}=\dfrac{1}{x^3 + {y}^{3} + {z}^{3} - 3xyz} \dfrac{\partial }{\partial z}(x^3 + {y}^{3} + {z}^{3} - 3xyz)$

$\rm\implies \:\rm \: \dfrac{\partial u}{\partial z}=\dfrac{ {3z}^{2} - 3xy}{x^3 + {y}^{3} + {z}^{3} - 3xyz} - - - (3)$

On adding equation (1), (2) and (3), we get

$\rm \: \dfrac{\partial u}{\partial x} + \dfrac{\partial u}{\partial y} + \dfrac{\partial u}{\partial z}$

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

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

$\rm \: = \: \dfrac{3}{x + y + z}$

$\rm\implies \:\bf\: \dfrac{\partial u}{\partial x} + \dfrac{\partial u}{\partial y} + \dfrac{\partial u}{\partial z} = \frac{3}{x + y + z} - - - (4)$

Now, Consider

$\rm \bigg( \dfrac{ \partial}{ \partial x} + \dfrac{ \partial}{ \partial y} + \dfrac{ \partial}{ \partial z} \bigg) ^2u$

can be rewritten as

$\rm \: = \: \bigg(\dfrac{\partial }{\partial x} + \dfrac{\partial }{\partial y} + \dfrac{\partial }{\partial z}\bigg)\bigg(\dfrac{\partial }{\partial x} + \dfrac{\partial }{\partial y} + \dfrac{\partial }{\partial z}\bigg)u$

$\rm \: = \: \bigg(\dfrac{\partial }{\partial x} + \dfrac{\partial }{\partial y} + \dfrac{\partial }{\partial z}\bigg)\bigg(\dfrac{\partial u}{\partial x} + \dfrac{\partial u}{\partial y} + \dfrac{\partial u}{\partial z}\bigg)$

$\rm \: = \: \bigg(\dfrac{\partial }{\partial x} + \dfrac{\partial }{\partial y} + \dfrac{\partial }{\partial z}\bigg)\dfrac{3}{x + y + z}$

$\rm \: = \: \dfrac{\partial }{\partial x} \dfrac{3}{x + y + z} + \dfrac{\partial }{\partial y} \dfrac{3}{x + y + z} + \dfrac{\partial }{\partial z} \dfrac{3}{x + y + z}$

$\rm \: = \: \dfrac{ - 3}{(x + y + z)^{2} } + \dfrac{ - 3}{(x + y + z)^{2} } - \dfrac{ - 3}{(x + y + z)^{2} }$

$\rm \: = \: \dfrac{ - 9}{(x + y + z)^{2} }$

Hence,

$\rm\implies \boxed{ \bf{ \:\bigg( \frac{ \partial}{ \partial x} + \frac{ \partial}{ \partial y} + \frac{ \partial}{ \partial z} \bigg) ^2u = \frac{ - 9}{(x + y + z) {}^{2} } \: }}$

$\rule{190pt}{2pt}$

Formulae Used :-

$\boxed{ \rm{ \:\dfrac{\partial }{\partial x}logx = \frac{1}{x} \: }}$

$\boxed{ \rm{ \:\dfrac{\partial }{\partial x} {x}^{n} = {nx}^{n - 1} \: }}$

$\boxed{ \rm{ \:\dfrac{\partial }{\partial x} k = 0\: }}$

$\boxed{ \rm{ \:\dfrac{\partial }{\partial x} \frac{1}{ {x}^{n} } = \frac{ - n}{ {x}^{n + 1} } \: }}$