SSC (English Medium) Class 8Maharashtra State Board
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# Simplify : a 2 + 10 a + 21 a 2 + 6 a − 7 × a 2 − 1 a + 3 - SSC (English Medium) Class 8 - Mathematics

#### Question

Simplify :

$\frac{a^2 + 10a + 21}{a^2 + 6a - 7} \times \frac{a^2 - 1}{a + 3}$

#### Solution

It is known that,

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

$\ \frac{a^2 + 10a + 21}{a^2 + 6a - 7} \times \frac{a^2 - 1}{a + 3}$

$= \frac{a^2 + 7a + 3a + 21}{a^2 + 7a - a - 7} \times \frac{\left( a + 1 \right)\left( a - 1 \right)}{\left( a + 3 \right)}$

$= \frac{a\left( a + 7 \right) + 3\left( a + 7 \right)}{a \left( a + 7 \right) - 1\left( a + 7 \right)} \times \frac{\left( a + 1 \right)\left( a - 1 \right)}{\left( a + 3 \right)}$

$= \frac{\left( a + 7 \right)\left( a + 3 \right)}{\left( a - 1 \right)\left( a + 7 \right)} \times \frac{\left( a + 1 \right)\left( a - 1 \right)}{\left( a + 3 \right)}$

$= \left( a + 1 \right)$

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: 2 | Page no. 33
Solution Simplify : a 2 + 10 a + 21 a 2 + 6 a − 7 × a 2 − 1 a + 3 Concept: Factors of A3 - B3.
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