Question

# Simplify: $\dfrac{{{a^2} - 16}}{{{a^3} - 8}} \times \dfrac{{2{a^2} - 3a - 2}}{{2{a^2} + 9a + 4}} \div \dfrac{{3{a^2} - 11a - 4}}{{{a^2} + 2a + 4}}$

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.