SSC (English Medium) Class 8Maharashtra State Board
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# Simplify : a 3 − 27 5 a 2 − 16 a + 3 ÷ a 2 + 3 a + 9 25 a 2 − 1 - SSC (English Medium) Class 8 - Mathematics

#### Question

Simplify :

$\frac{a^3 - 27}{5 a^2 - 16a + 3} \div \frac{a^2 + 3a + 9}{25 a^2 - 1}$

#### Solution

It is known that,

$a^2 - b^2 = \left( a + b \right)\left( a - b \right) ; a^3 - b^3 = \left( a - b \right)\left( a^2 + ab + b^2 \right)$

$\ \frac{a^3 - 27}{5 a^2 - 16a + 3} \div \frac{a^2 + 3a + 9}{25 a^2 - 1}$

$= \frac{\left( a \right)^3 - \left( 3 \right)^3}{5 a^2 - 15a - a + 3} \div \frac{a^2 + 3a + 9}{\left( 5a \right)^2 - \left( 1 \right)^2}$

$= \frac{\left( a - 3 \right)\left\{ \left( a \right)^2 + \left( a \right) \times \left( 3 \right) + \left( 3 \right)^2 \right\}}{5a\left( a - 3 \right) - 1\left( a - 3 \right)} \div \frac{a^2 + 3a + 9}{\left( 5a + 1 \right)\left( 5a - 1 \right)}$

$= \frac{\left( a - 3 \right)\left( a^2 + 3a + 9 \right)}{\left( 5a - 1 \right)\left( a - 3 \right)} \div \frac{\left( a^2 + 3a + 9 \right)}{\left( 5a + 1 \right)\left( 5a - 1 \right)}$

$= \frac{\left( a - 3 \right)\left( a^2 + 3a + 9 \right)}{\left( 5a - 1 \right)\left( a - 3 \right)} \times \frac{\left( 5a + 1 \right)\left( 5a - 1 \right)}{\left( a^2 + 3a + 9 \right)}$

$= 5a + 1$

Is there an error in this question or solution?

#### APPEARS IN

Balbharati Solution for Balbharati Class 8 Mathematics (2019 to Current)
Chapter 6: Factorisation of Algebraic expressions
Practice Set 6.4 | Q: 7 | Page no. 33
Solution Simplify : a 3 − 27 5 a 2 − 16 a + 3 ÷ a 2 + 3 a + 9 25 a 2 − 1 Concept: Factors of A3 - B3.
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