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Factorise: 8p3 − 27 P 3 - SSC (English Medium) Class 8 - Mathematics

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Question

Factorise:

8p−\[\frac{27}{p^3}\]

Solution

It is known that,

\[a^3  -  b^3    =   \left( a - b \right)\left( a^2 + b^2 + ab \right)\]

\[\  8 p^3  - \frac{27}{p^3}\] 

\[ =  \left( 2p \right)^3  -  \left( \frac{3}{p} \right)^3 \] 

\[ = \left( 2p - \frac{3}{p} \right)\left\{ \left( 2p \right)^2 + \left( \frac{3}{p} \right)^2 + \left( 2p \right) \times \left( \frac{3}{p} \right) \right\}\] 

\[ = \left( 2p - \frac{3}{p} \right)\left( 4 p^2 + \frac{9}{p^2} + 6 \right)\]

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APPEARS IN

 Balbharati Solution for Balbharati Class 8 Mathematics (2019 to Current)
Chapter 6: Factorisation of Algebraic expressions
Practice Set 6.3 | Q: 1.5 | Page no. 32
Solution Factorise: 8p3 − 27 P 3 Concept: Factors of A3 - B3.
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