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Factorise: 3a7 – 192ab6 - Mathematics

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Question

Factorise:

3a7 – 192ab6

Sum
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Solution

Given expression: 3a7 – 192ab6 

⇒ 3a(a2)3 – (4b2)3]

⇒ 3a[(a2)3 – (4b2)3]

We know that,

a3 – b3 = (a – b) (a2 + ab + b2)

Hence,

⇒ 3a(a2 – 4b2) (a4 + 4a2b2 + 16b4)

⇒ 3a[(a)2 – (2b)2 ] (a4 + 4a2b2 + 16b4)

⇒ 3a(a – 2b) (a + 2b) (a4 + 4a2b2 + 16b4)

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Chapter 4: Factorisation - EXERCISE 4D [Page 46]

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B Nirmala Shastry Mathematics [English] Class 9 ICSE
Chapter 4 Factorisation
EXERCISE 4D | Q I. 16. | Page 46
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