Answer
427.5k+ 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.
Complete step-by-step 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.
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.
Complete step-by-step 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.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)