Answer
454.2k+ views
Hint: For simplification, use
$
\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right) \\
\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right) \\
$
Given equation is
\[\dfrac{{{a^2} - 16}}{{{a^3} - 8}} \times \dfrac{{2{a^2} - 3a - 2}}{{2{a^2} + 9a + 4}} \div \dfrac{{3{a^2} - 11a - 4}}{{{a^2} + 2a + 4}}..................\left( 1 \right)\]
Now, simplify the terms in the expression using the formulas given in the hint,
$
\Rightarrow \left( {{a^2} - 16} \right) = \left( {{a^2} - {4^2}} \right) = \left( {a - 4} \right)\left( {a + 4} \right) \\
\Rightarrow \left( {{a^3} - 8} \right) = \left( {{a^3} - {2^3}} \right) = \left( {a - 2} \right)\left( {{a^2} + 4 + 2a} \right) \\
$
Now, factorize the remaining terms of the expression
$
\Rightarrow \left( {2{a^2} - 3a - 2} \right) = \left( {2{a^2} - 4a + a - 2} \right) = 2a\left( {a - 2} \right) + 1\left( {a - 2} \right) = \left( {2a + 1} \right)\left( {a - 2} \right) \\
\Rightarrow \left( {2{a^2} + 9a + 4} \right) = \left( {2{a^2} + 8a + a + 4} \right) = 2a\left( {a + 4} \right) + 1\left( {a + 4} \right) = \left( {2a + 1} \right)\left( {a + 4} \right) \\
\Rightarrow \left( {3{a^2} - 11a - 4} \right) = \left( {3{a^2} - 12a + a - 4} \right) = 3a\left( {a - 4} \right) + 1\left( {a - 4} \right) = \left( {3a + 1} \right)\left( {a - 4} \right) \\
$
Substitute those in equation 1
$
\Rightarrow \dfrac{{{a^2} - 16}}{{{a^3} - 8}} \times \dfrac{{2{a^2} - 3a - 2}}{{2{a^2} + 9a + 4}} \div \dfrac{{3{a^2} - 11a - 4}}{{{a^2} + 2a + 4}} \\
\Rightarrow \dfrac{{\left( {a - 4} \right)\left( {a + 4} \right)}}{{\left( {a - 2} \right)\left( {{a^2} + 4 + 2a} \right)}} \times \dfrac{{\left( {2a + 1} \right)\left( {a - 2} \right)}}{{\left( {2a + 1} \right)\left( {a + 4} \right)}} \div \dfrac{{\left( {3a + 1} \right)\left( {a - 4} \right)}}{{\left( {{a^2} + 4 + 2a} \right)}}...........\left( 2 \right) \\
$
Now we know if we convert division into multiplication, then numerator and denominator will interchange, therefore equation 2 can be written as
\[\Rightarrow \dfrac{{\left( {a - 4} \right)\left( {a + 4} \right)}}{{\left( {a - 2} \right)\left( {{a^2} + 4 + 2a} \right)}} \times \dfrac{{\left( {2a + 1} \right)\left( {a - 2} \right)}}{{\left( {2a + 1} \right)\left( {a + 4} \right)}} \times \dfrac{{\left( {{a^2} + 4 + 2a} \right)}}{{\left( {3a + 1} \right)\left( {a - 4} \right)}} \\
\]
Now as we see all terms are cancel out only one term is remaining which is \[\dfrac{1}{{\left( {3a + 1} \right)}}\]
\[ \Rightarrow \dfrac{{{a^2} - 16}}{{{a^3} - 8}} \times \dfrac{{2{a^2} - 3a - 2}}{{2{a^2} + 9a + 4}} \div \dfrac{{3{a^2} - 11a - 4}}{{{a^2} + 2a + 4}} = \dfrac{1}{{\left( {3a + 1} \right)}}\]
So, this is the required simplification.
Note: - In these types of questions the key concept is to use the formula of $\left( {{a^2} - {b^2}} \right)$ and $\left( {{a^3} - {b^3}} \right)$ to simplify the given expression to get the required result.
$
\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right) \\
\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right) \\
$
Given equation is
\[\dfrac{{{a^2} - 16}}{{{a^3} - 8}} \times \dfrac{{2{a^2} - 3a - 2}}{{2{a^2} + 9a + 4}} \div \dfrac{{3{a^2} - 11a - 4}}{{{a^2} + 2a + 4}}..................\left( 1 \right)\]
Now, simplify the terms in the expression using the formulas given in the hint,
$
\Rightarrow \left( {{a^2} - 16} \right) = \left( {{a^2} - {4^2}} \right) = \left( {a - 4} \right)\left( {a + 4} \right) \\
\Rightarrow \left( {{a^3} - 8} \right) = \left( {{a^3} - {2^3}} \right) = \left( {a - 2} \right)\left( {{a^2} + 4 + 2a} \right) \\
$
Now, factorize the remaining terms of the expression
$
\Rightarrow \left( {2{a^2} - 3a - 2} \right) = \left( {2{a^2} - 4a + a - 2} \right) = 2a\left( {a - 2} \right) + 1\left( {a - 2} \right) = \left( {2a + 1} \right)\left( {a - 2} \right) \\
\Rightarrow \left( {2{a^2} + 9a + 4} \right) = \left( {2{a^2} + 8a + a + 4} \right) = 2a\left( {a + 4} \right) + 1\left( {a + 4} \right) = \left( {2a + 1} \right)\left( {a + 4} \right) \\
\Rightarrow \left( {3{a^2} - 11a - 4} \right) = \left( {3{a^2} - 12a + a - 4} \right) = 3a\left( {a - 4} \right) + 1\left( {a - 4} \right) = \left( {3a + 1} \right)\left( {a - 4} \right) \\
$
Substitute those in equation 1
$
\Rightarrow \dfrac{{{a^2} - 16}}{{{a^3} - 8}} \times \dfrac{{2{a^2} - 3a - 2}}{{2{a^2} + 9a + 4}} \div \dfrac{{3{a^2} - 11a - 4}}{{{a^2} + 2a + 4}} \\
\Rightarrow \dfrac{{\left( {a - 4} \right)\left( {a + 4} \right)}}{{\left( {a - 2} \right)\left( {{a^2} + 4 + 2a} \right)}} \times \dfrac{{\left( {2a + 1} \right)\left( {a - 2} \right)}}{{\left( {2a + 1} \right)\left( {a + 4} \right)}} \div \dfrac{{\left( {3a + 1} \right)\left( {a - 4} \right)}}{{\left( {{a^2} + 4 + 2a} \right)}}...........\left( 2 \right) \\
$
Now we know if we convert division into multiplication, then numerator and denominator will interchange, therefore equation 2 can be written as
\[\Rightarrow \dfrac{{\left( {a - 4} \right)\left( {a + 4} \right)}}{{\left( {a - 2} \right)\left( {{a^2} + 4 + 2a} \right)}} \times \dfrac{{\left( {2a + 1} \right)\left( {a - 2} \right)}}{{\left( {2a + 1} \right)\left( {a + 4} \right)}} \times \dfrac{{\left( {{a^2} + 4 + 2a} \right)}}{{\left( {3a + 1} \right)\left( {a - 4} \right)}} \\
\]
Now as we see all terms are cancel out only one term is remaining which is \[\dfrac{1}{{\left( {3a + 1} \right)}}\]
\[ \Rightarrow \dfrac{{{a^2} - 16}}{{{a^3} - 8}} \times \dfrac{{2{a^2} - 3a - 2}}{{2{a^2} + 9a + 4}} \div \dfrac{{3{a^2} - 11a - 4}}{{{a^2} + 2a + 4}} = \dfrac{1}{{\left( {3a + 1} \right)}}\]
So, this is the required simplification.
Note: - In these types of questions the key concept is to use the formula of $\left( {{a^2} - {b^2}} \right)$ and $\left( {{a^3} - {b^3}} \right)$ to simplify the given expression to get the required result.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 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)
Give 10 examples for herbs , shrubs , climbers , creepers
![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)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english 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)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)