Question

Evaluate $u=\log \left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz \right)\Rightarrow \left( x+y+z \right)\left( {{u}_{z}}+{{u}_{y}}+{{u}_{z}} \right)=$
A. 0
B. $x-y+z$
C. 2
D. 3

Answer 1

${{u}_{x}},{{u}_{y}}\ and\ {{u}_{z}}$ are standard notations used for partial differentiation of u with respect to x, y and z respectively. First find out ${{u}_{x}},{{u}_{y}}\ and\ {{u}_{z}}$ by partially differentiating the equation $''u=\log \left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz \right)''$ with respect to x, y and z respectively. Then put the values obtained in $\left( x+y+z \right)\left( {{u}_{z}}+{{u}_{y}}+{{u}_{z}} \right)$ and further simplify to get the answer. Differentiation of $\log x$ with respect to x is $\dfrac{1}{x}$.

Complete step-by-step answer:
Given: $u=\log \left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz \right)$ and we have to find $\left( x+y+z \right)\left( {{u}_{z}}+{{u}_{y}}+{{u}_{z}} \right)$.
Let us first find ${{u}_{x}},{{u}_{y}}\ and\ {{u}_{z}}$
${{u}_{x}}=$ Partial differentiation of u with respect to x denoted by $''\dfrac{\partial u}{\partial x}''$.
In calculating partial differentiation with respect to any variable, we assume other variables to be constant.
So, for calculating $\dfrac{\partial u}{\partial x}$, we will assume ‘y’ and ‘z’ to be constants. So, u will be only a function of x:
$u=\log \left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz \right)$
We know differentiation of $\log x$ is $\dfrac{1}{x}$.
Now, using Product rule of differentiation,
$\begin{aligned} & {{u}_{x}}=\dfrac{\partial u}{\partial x}=\dfrac{\partial }{\partial \left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz \right)}\log \left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz \right)\times \dfrac{\partial }{\partial x}\left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz \right) \\\ & =\left( \dfrac{1}{\left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz \right)} \right)\times \left( 3{{x}^{2}}+0+0-3yz \right) \\\ & \left[ \text{As we have assume y }\\!\\!\And\\!\\!\text{ z to be constants} \right] \\\ & ={{u}_{x}}=\dfrac{\partial u}{\partial x}=\dfrac{3{{x}^{2}}-3yz}{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz}............\left( 1 \right) \\\ \end{aligned}$
Similarly, assuming ‘x’ and ‘z’ to be constants and by using the Product rule of differentiation, we will calculate $\dfrac{\partial u}{\partial y}$.
$\begin{aligned} & {{u}_{y}}=\dfrac{\partial u}{\partial y}=\dfrac{\partial }{\partial \left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz \right)}\log \left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz \right)\times \dfrac{\partial }{\partial y}\left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz \right) \\\ & =\dfrac{1}{\left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz \right)}\times \left( 0+3{{y}^{2}}+0-3xz \right) \\\ & \left[ \text{As we have assume x }\\!\\!\And\\!\\!\text{ z to be constants} \right] \\\ & ={{u}_{y}}=\dfrac{\partial u}{\partial y}=\dfrac{3{{y}^{2}}-3xz}{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz}............\left( 2 \right) \\\ \end{aligned}$
Similarly, assuming ‘x’ and ‘y’ to be constants and by using Product rule of differentiation, we will get,
$\begin{aligned} & {{u}_{z}}=\dfrac{\partial u}{\partial z}=\dfrac{\partial }{\partial \left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz \right)}\log \left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz \right)\times \dfrac{\partial }{\partial z}\left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz \right) \\\ & =\dfrac{1}{\left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz \right)}\times \left( 0+0+3{{z}^{2}}-3xy \right) \\\ & \left[ \text{As we have assume x }\\!\\!\And\\!\\!\text{ y to be constants} \right] \\\ & ={{u}_{z}}=\dfrac{\partial u}{\partial z}=\dfrac{3{{z}^{2}}-3xy}{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz}............\left( 3 \right) \\\ \end{aligned}$
According to the question, we have to find $\left( x+y+z \right)\left( {{u}_{z}}+{{u}_{y}}+{{u}_{z}} \right)$.
On putting values of ${{u}_{x}},{{u}_{y}}\ and\ {{u}_{z}}$ from equation (1), (2) and (3) respectively, we will get,
$\left( x+y+z \right)\left( {{u}_{z}}+{{u}_{y}}+{{u}_{z}} \right)=\left( x+y+z \right)\left[ \dfrac{3{{x}^{2}}-3yz}{\left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz \right)}+\dfrac{3{{y}^{2}}-3xz}{\left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz \right)}+\dfrac{3{{z}^{2}}-3xy}{\left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz \right)} \right]$Taking LCM and adding, we will get,
$\left( x+y+z \right)\left( {{u}_{z}}+{{u}_{y}}+{{u}_{z}} \right)=\left( x+y+z \right)\left[ \dfrac{\left( 3{{x}^{2}}-3yz \right)+\left( 3{{y}^{2}}-3xz \right)+\left( 3{{z}^{2}}-3xy \right)}{\left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz \right)} \right]$
We know the identity:
$\left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz \right)=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)$
Using this identity in above equation, we will get,
$\left( x+y+z \right)\left( {{u}_{z}}+{{u}_{y}}+{{u}_{z}} \right)=\left( x+y+z \right)\left[ \dfrac{\left( 3{{x}^{2}}-3yz \right)+\left( 3{{y}^{2}}-3xz \right)+\left( 3{{z}^{2}}-3xy \right)}{\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)} \right]$
On the RHS, dividing both numerator and denominator by $\left( x+y+z \right)$, we will get,
$\left( x+y+z \right)\left( {{u}_{z}}+{{u}_{y}}+{{u}_{z}} \right)=\dfrac{\left( 3{{x}^{2}}-3yz \right)+\left( 3{{y}^{2}}-3xz \right)+\left( 3{{z}^{2}}-3xy \right)}{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)}$
Rearranging the terms in the numerator of RHS, we will get,
$\left( x+y+z \right)\left( {{u}_{z}}+{{u}_{y}}+{{u}_{z}} \right)=\dfrac{\left( 3{{x}^{2}}+3{{y}^{2}}+3{{z}^{2}}-3xy-3yz-3zx \right)}{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)}$
Taking 3 common from the terms in numerator of RHS, we will get,
$\left( x+y+z \right)\left( {{u}_{z}}+{{u}_{y}}+{{u}_{z}} \right)=\dfrac{3\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)}{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)}$
Now, dividing both numerator and denominator by $\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)$ on RHS, we will get,
$\left( x+y+z \right)\left( {{u}_{z}}+{{u}_{y}}+{{u}_{z}} \right)=3$
Hence, the required value of $\left( x+y+z \right)\left( {{u}_{z}}+{{u}_{y}}+{{u}_{z}} \right)$ is 3 and option (D) is the correct answer.

Note: We have used an identity $\left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz \right)=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)$. If you don’t know this identity then, we can prove this identity as follows:
$LHS={{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz$
On adding and subtracting $\left( 3{{x}^{2}}y+3x{{y}^{2}} \right)$ we will get,
$LHS={{x}^{3}}+{{y}^{3}}+3{{x}^{2}}y+3x{{y}^{2}}+{{z}^{3}}-3xyz-3{{x}^{2}}y-3x{{y}^{2}}$
We know,
$\begin{aligned} & {{\left( x+y \right)}^{3}}={{x}^{3}}+{{y}^{3}}+3xy\left( x+y \right) \\\ & ={{x}^{3}}+{{y}^{3}}+3{{x}^{2}}y+3x{{y}^{2}} \\\ \end{aligned}$
So, replacing ${{x}^{3}}+{{y}^{3}}+3{{x}^{2}}y+3x{{y}^{2}}$ by ${{\left( x+y \right)}^{3}}$, we will get,
$LHS={{\left( x+y \right)}^{3}}+{{z}^{3}}-3xyz-3{{x}^{2}}y-3x{{y}^{2}}$
Taking 3xy common from the last three terms, we will get,
$LHS={{\left( x+y \right)}^{3}}+{{z}^{3}}-3xy(x+y+z)$
We know ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$
Let $a=x+y\ and\ b=z$
$\Rightarrow {{\left( x+y \right)}^{3}}+{{z}^{3}}=\left( x+y+z \right)\left[ {{\left( x+y \right)}^{2}}-\left( x+y \right)z+{{z}^{2}} \right]$
Replacing ${{\left( x+y \right)}^{3}}+{{z}^{3}}$ by $\left( x+y+z \right)\left[ {{\left( x+y \right)}^{2}}-\left( x+y \right)z+{{z}^{2}} \right]$ we will get,
$LHS=\left( x+y+z \right)\left[ {{\left( x+y \right)}^{2}}-\left( x+y \right)z+{{z}^{2}} \right]-3xy(x+y+z)$
Taking $\left( x+y+z \right)$, we will get,
$LHS=\left( x+y+z \right)\left[ {{\left( x+y \right)}^{2}}-\left( x+y \right)z+{{z}^{2}}-3xy \right]$
Replacing ${{\left( x+y \right)}^{2}}$ with ${{x}^{2}}+{{y}^{2}}+2xy$, we will get,

& LHS=\left( x+y+z \right)\left[ {{x}^{2}}+{{y}^{2}}+2xy-xz-yz+{{z}^{2}}-3xy \right] \\\ & LHS=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xz-yz-zx \right) \\\ & LHS=RHS \\\ \end{aligned}$$ Identity Proved.