Maharashtra State BoardSSC (English Medium) 8th Standard
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Factorise: 16a3 − 128 B 3 - Mathematics

Sum

Factorise:

16a−\[\frac{128}{b^3}\]

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Solution

It is known that,

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

\[\  16 a^3  - \frac{128}{b^3}\] 

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

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

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

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

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APPEARS IN

Balbharati Mathematics 8th Standard Maharashtra State Board
Chapter 6 Factorisation of Algebraic expressions
Practice Set 6.3 | Q 1.8 | Page 32
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