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Factorise: 16a3 − 128 B 3 - SSC (English Medium) Class 8 - Mathematics

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Question

Factorise:

16a−\[\frac{128}{b^3}\]

Solution

It is known that,

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

\[\  16 a^3  - \frac{128}{b^3}\] 

\[ = 16\left[ a^3 - \frac{8}{b^3} \right]\] 

\[ = 16\left[ \left( a \right)^3 - \left( \frac{2}{b} \right)^3 \right]\] 

\[ = 16\left[ \left( a - \frac{2}{b} \right)\left\{ \left( a \right)^2 + \left( \frac{2}{b} \right)^2 + \left( a \right) \times \left( \frac{2}{b} \right) \right\} \right]\] 

\[ = 16\left( a - \frac{2}{b} \right)\left( a^2 + \frac{4}{b^2} + \frac{2a}{b} \right)\]

  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 6.3 | Q: 1.8 | Page no. 32
Solution Factorise: 16a3 − 128 B 3 Concept: Factors of A3 - B3.
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