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Simplify: 4^3โข๐‘› + 4^2โข๐‘› โ‹… 4^๐‘›+1/(4 ร— 4^๐‘›)^3 โˆ’ 4 ร— 4^3โข๐‘› - Mathematics

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Question

Simplify:

`(4^(3n) + 4^(2n) * 4^(n + 1))/((4 xx 4^n)^3 - 4 xx 4^(3n))`

Simplify
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Solution

`(4^(3n) + 4^(2n) * 4^(n + 1))/((4 xx 4^n)^3 - 4 xx 4^(3n))`

= `(4^(3n) + 4^(3n + 1))/(4^(3(n + 1)) - 4^(3n + 1))`

= `(4^(3n)(1 + 4))/(4^(3n + 1)(4^2 - 1))`

= `(5 * 4^(3n))/(4^(3n + 1) * 15)`

= `5/(15 * 4)`

= `1/12`

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Chapter 6: Indices - MISCELLANEOUS EXERCISE [Page 69]

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B Nirmala Shastry Mathematics [English] Class 9 ICSE
Chapter 6 Indices
MISCELLANEOUS EXERCISE | Q 3. (i) | Page 69
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