Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The value of $\dfrac{{\left( {\dfrac{x}{y} - \dfrac{y}{x}} \right)\left( {\dfrac{y}{z} - \dfrac{z}{y}} \right)\left( {\dfrac{z}{x} - \dfrac{x}{z}} \right)}}{{\left( {\dfrac{1}{{{x^2}}} - \dfrac{1}{{{y^2}}}} \right)\left( {\dfrac{1}{{{y^2}}} - \dfrac{1}{{{z^2}}}} \right)\left( {\dfrac{1}{{{z^2}}} - \dfrac{1}{{{x^2}}}} \right)}}$ isA. $- {x^2}{y^2}{z^2}$B. ${x^2}{y^2}{z^2}$C. $1$D. $xyz$

Last updated date: 13th Jun 2024
Total views: 402.6k
Views today: 11.02k
Verified
402.6k+ views
Hint: Solve the denominator and numerator separately to avoid the complex calculations. Then put their respective simplified values of denominator and numerator in the given expression.
Rule of division for dividing rational expressions: $\dfrac{{\left( {\dfrac{a}{b}} \right)}}{{\left( {\dfrac{c}{d}} \right)}} = \left( {\dfrac{a}{b}} \right) \times \left( {\dfrac{c}{d}} \right)$
Use this rule and then cancel the common terms to get the answer.

Step-1
Let $\dfrac{{\left( {\dfrac{x}{y} - \dfrac{y}{x}} \right)\left( {\dfrac{y}{z} - \dfrac{z}{y}} \right)\left( {\dfrac{z}{x} - \dfrac{x}{z}} \right)}}{{\left( {\dfrac{1}{{{x^2}}} - \dfrac{1}{{{y^2}}}} \right)\left( {\dfrac{1}{{{y^2}}} - \dfrac{1}{{{z^2}}}} \right)\left( {\dfrac{1}{{{z^2}}} - \dfrac{1}{{{x^2}}}} \right)}} = \dfrac{A}{B}$ … $\left( 1 \right)$
$\Rightarrow A = \left( {\dfrac{x}{y} - \dfrac{y}{x}} \right)\left( {\dfrac{y}{z} - \dfrac{z}{y}} \right)\left( {\dfrac{z}{x} - \dfrac{x}{z}} \right)$ and $B = \left( {\dfrac{1}{{{x^2}}} - \dfrac{1}{{{y^2}}}} \right)\left( {\dfrac{1}{{{y^2}}} - \dfrac{1}{{{z^2}}}} \right)\left( {\dfrac{1}{{{z^2}}} - \dfrac{1}{{{x^2}}}} \right)$
Solve the value of A and B separately.
Step-2
Solve for A
$\Rightarrow A = \left( {\dfrac{x}{y} - \dfrac{y}{x}} \right)\left( {\dfrac{y}{z} - \dfrac{z}{y}} \right)\left( {\dfrac{z}{x} - \dfrac{x}{z}} \right)$
Simplify the terms in brackets by taking LCM
On simplifying
$\Rightarrow A = \left( {\dfrac{{{x^2} - {y^2}}}{{xy}}} \right)\left( {\dfrac{{{y^2} - {z^2}}}{{zy}}} \right)\left( {\dfrac{{{z^2} - {x^2}}}{{xz}}} \right)$
Multiply the terms of denominator
$\Rightarrow A = \dfrac{{\left( {{x^2} - {y^2}} \right)\left( {{y^2} - {z^2}} \right)\left( {{z^2} - {x^2}} \right)}}{{{x^2}{y^2}{z^2}}}$
Step-3
Now solve for B
$B = \left( {\dfrac{1}{{{x^2}}} - \dfrac{1}{{{y^2}}}} \right)\left( {\dfrac{1}{{{y^2}}} - \dfrac{1}{{{z^2}}}} \right)\left( {\dfrac{1}{{{z^2}}} - \dfrac{1}{{{x^2}}}} \right)$
Simplify the terms in brackets by taking LCM

On simplifying we get
$\Rightarrow B = \left( {\dfrac{{{y^2} - {x^2}}}{{{x^2}{y^2}}}} \right)\left( {\dfrac{{{z^2} - {y^2}}}{{{y^2}{z^2}}}} \right)\left( {\dfrac{{{x^2} - {z^2}}}{{{z^2}{x^2}}}} \right)$
Now Multiply the terms of denominator

$\Rightarrow B = \dfrac{{\left( {{y^2} - {x^2}} \right)\left( {{z^2} - {y^2}} \right)\left( {{x^2} - {z^2}} \right)}}{{{x^4}{y^4}{z^4}}}$
Step- 4

Now from the step-2 and step-3 substitute the value of $A$and $B$ in $\left( 1 \right)$
We get the whole expression as
$\dfrac{A}{B} = \dfrac{{\dfrac{{\left( {{x^2} - {y^2}} \right)\left( {{y^2} - {z^2}} \right)\left( {{z^2} - {x^2}} \right)}}{{{x^2}{y^2}{z^2}}}}}{{\dfrac{{\left( {{y^2} - {x^2}} \right)\left( {{z^2} - {y^2}} \right)\left( {{x^2} - {z^2}} \right)}}{{{x^4}{y^4}{z^4}}}}}$
We can rewrite the above expression as
$\Rightarrow \dfrac{A}{B} = \dfrac{{\dfrac{{\left( {{x^2} - {y^2}} \right)\left( {{y^2} - {z^2}} \right)\left( {{z^2} - {x^2}} \right)}}{{{x^2}{y^2}{z^2}}}}}{{\dfrac{{\left( {{y^2} - {x^2}} \right)\left( {{z^2} - {y^2}} \right)\left( {{x^2} - {z^2}} \right)}}{{\left( {{x^2}{y^2}{z^2}} \right)\left( {{x^2}{y^2}{z^2}} \right)}}}}$
Take the – sign common from the numerator of $B$(make the expression of $B$similar to $A$) in order to cancel the common terms
$\Rightarrow \dfrac{A}{B} = \dfrac{{\dfrac{{\left( {{x^2} - {y^2}} \right)\left( {{y^2} - {z^2}} \right)\left( {{z^2} - {x^2}} \right)}}{{{x^2}{y^2}{z^2}}}}}{{\dfrac{{ - \left( {{x^2} - {y^2}} \right)\left( {{y^2} - {z^2}} \right)\left( {{z^2} - {x^2}} \right)}}{{\left( {{x^2}{y^2}{z^2}} \right)\left( {{x^2}{y^2}{z^2}} \right)}}}}$
Now cancel the common terms
$\Rightarrow \dfrac{A}{B} = \dfrac{{\dfrac{1}{1}}}{{\dfrac{{ - 1}}{{\left( {{x^2}{y^2}{z^2}} \right)}}}}$
$\Rightarrow \dfrac{A}{B} = \dfrac{1}{{\dfrac{{ - 1}}{{\left( {{x^2}{y^2}{z^2}} \right)}}}}$ $\because \dfrac{1}{1} = 1$ … $\left( 2 \right)$
Rule of division for dividing rational expressions: $\dfrac{{\left( {\dfrac{a}{b}} \right)}}{{\left( {\dfrac{c}{d}} \right)}} = \left( {\dfrac{a}{b}} \right) \times \left( {\dfrac{c}{d}} \right)$
By using above mentioned rule, $\left( 2 \right)$becomes
$\Rightarrow \dfrac{A}{B} = - \left( {{x^2}{y^2}{z^2}} \right)$
$\Rightarrow \dfrac{A}{B} = - {x^2}{y^2}{z^2}$
Hence the value of $\dfrac{{\left( {\dfrac{x}{y} - \dfrac{y}{x}} \right)\left( {\dfrac{y}{z} - \dfrac{z}{y}} \right)\left( {\dfrac{z}{x} - \dfrac{x}{z}} \right)}}{{\left( {\dfrac{1}{{{x^2}}} - \dfrac{1}{{{y^2}}}} \right)\left( {\dfrac{1}{{{y^2}}} - \dfrac{1}{{{z^2}}}} \right)\left( {\dfrac{1}{{{z^2}}} - \dfrac{1}{{{x^2}}}} \right)}}$ is $- {x^2}{y^2}{z^2}$
Hence, Option(A) is the correct answer.

Note: Do not solve the whole fraction together. In these types of questions, solve the numerator and denominator part separately. while solving the parts(i.e. denominator and numerator) observe the question and do not multiply or solve the terms which may cancel later on. By doing so you can save your time as well as complexity of the question.