Question

factorise the following expression using suitable identity. 1+64a^3 please I need it fast really fast​

Answer 1

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1−64a3−12a+48a2

=(1)3−(4a)3−3(1)2(4a)+3(1)(4a)2

 Suitable identities is x3−y3−3x2y+3xy2=(x−y)3

∴(1)3−(4a)3−3(1)2(4a)+3(1)(4a)2

=(1−4a)3

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Answer 2

Answer:

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