Question

Find the value of the expression given below,
$\dfrac{{{a}^{3}}-27}{5{{a}^{2}}-16a+3}\div \dfrac{{{a}^{2}}+3a+9}{25{{a}^{2}}-1}$
A.5a – 1
B.5a + 1
C.a + 1
D.a – 1

Answer 1

Hint:Use the formulae $\dfrac{a}{b}\div \dfrac{c}{d}=\dfrac{a}{b}\times \dfrac{d}{c}$, ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\times \left( {{a}^{2}}+ab+{{b}^{2}} \right)$ , and ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\times \left( a-b \right)$ in the given expression and then solve the quadratic expression $5{{a}^{2}}-16a+3$ present in the denominator by replacing -16a by (- 15a – a) and after simplification you will get the final answer.

Complete step by step answer:
To solve the given expression we will write it down first and assume it as ‘S’, therefore we will get,
$S=\dfrac{{{a}^{3}}-27}{5{{a}^{2}}-16a+3}\div \dfrac{{{a}^{2}}+3a+9}{25{{a}^{2}}-1}$ …………………………………………………….. (1)
To proceed further in the solution we should know how division can be converted to multiplication as shown below,
$\dfrac{a}{b}\div \dfrac{c}{d}=\dfrac{a}{b}\times \dfrac{d}{c}$
By using above formula we can write our equation as,
$\therefore S=\dfrac{{{a}^{3}}-27}{5{{a}^{2}}-16a+3}\times \dfrac{25{{a}^{2}}-1}{{{a}^{2}}+3a+9}$
We will arrange the above equation such that we can use the simple algebraic formulae after rearrangement, therefore we will get,
$\therefore S=\dfrac{{{a}^{3}}-{{3}^{3}}}{5{{a}^{2}}-16a+3}\times \dfrac{{{5}^{2}}\times {{a}^{2}}-1}{{{a}^{2}}+3a+9}$
$\therefore S=\dfrac{{{a}^{3}}-{{3}^{3}}}{5{{a}^{2}}-16a+3}\times \dfrac{{{\left( 5a \right)}^{2}}-{{1}^{2}}}{{{a}^{2}}+3a+9}$
Now to solve the above equation we should know the formulae given below,
Formulae:
${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\times \left( {{a}^{2}}+ab+{{b}^{2}} \right)$ , ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\times \left( a-b \right)$
If we use the above formulae in ‘S’ we will get,
$\therefore S=\dfrac{\left( a-3 \right)\times \left( {{a}^{2}}+3a+{{3}^{2}} \right)}{5{{a}^{2}}-16a+3}\times \dfrac{\left( 5a+1 \right)\times \left( 5a-1 \right)}{{{a}^{2}}+3a+9}$
Now if we observe the above equation carefully then we can easily cancel out $\left( {{a}^{2}}+3a+{{3}^{2}} \right)$ as it is present in both numerator and denominator, therefore by cancelling out $\left( {{a}^{2}}+3a+{{3}^{2}} \right)$ we will get,
$\therefore S=\dfrac{\left( a-3 \right)}{5{{a}^{2}}-16a+3}\times \left( 5a+1 \right)\times \left( 5a-1 \right)$
The expression $5{{a}^{2}}-16a+3$ present in the denominator of the above equation we can replace -16a by (- 15a – a) therefore by substituting it in the above equation we will get,
$\therefore S=\dfrac{\left( a-3 \right)}{5{{a}^{2}}-15a-a+3}\times \left( 5a+1 \right)\times \left( 5a-1 \right)$
Taking 5 common in the denominator of the above equation we will get,
$\therefore S=\dfrac{\left( a-3 \right)}{5a\left( a-3 \right)-a+3}\times \left( 5a+1 \right)\times \left( 5a-1 \right)$
Also taking -1 common from the denominator of the above equation we will get,
$\therefore S=\dfrac{\left( a-3 \right)}{5a\left( a-3 \right)-\left( a-3 \right)}\times \left( 5a+1 \right)\times \left( 5a-1 \right)$
Now, taking (a – 3) common from the denominator of the above equation we will get,
$\therefore S=\dfrac{\left( a-3 \right)}{\left( a-3 \right)\left( 5a-1 \right)}\times \left( 5a+1 \right)\times \left( 5a-1 \right)$
If we observe the above equation carefully then we will come to know that we can easily cancel out (a – 3) and (5a – 1) as they are both present in numerator as well as in denominator, therefore by cancelling them out we will get,
$\therefore S=\left( 5a+1 \right)$
If we put the value of ‘S’ from above equation in equation (1) we will get,
$\therefore \dfrac{{{a}^{3}}-27}{5{{a}^{2}}-16a+3}\div \dfrac{{{a}^{2}}+3a+9}{25{{a}^{2}}-1}=\left( 5a+1 \right)$
Therefore the value of the expression $\dfrac{{{a}^{3}}-27}{5{{a}^{2}}-16a+3}\div \dfrac{{{a}^{2}}+3a+9}{25{{a}^{2}}-1}$ is equal to (5a + 1).
Therefore the correct answer is option (b).

Note: Do not try to simplify the expressions like ${{a}^{2}}+3a+9$ which can’t be simplified easily because they can be cancelled by another simplified term like ${{a}^{3}}-27$ simplified as $\left[ \left( a-3 \right)\times \left( {{a}^{2}}+3a+{{3}^{2}} \right) \right]$ and you can easily cancel out ${{a}^{2}}+3a+9$.