Math, asked by aeki, 1 year ago

factorise-64a3+48a2-12a+1

Answers

Answered by Anonymous
229
   1 − 64a³ − 12a + 48a²
=(1)³ − (4a)³ − 3(1)²(4a) + 3(1)(4a)²
=(1 − 4a)²
=(1 − 4a)(1 − 4a)(1 − 4a)
Using the identity,a³b³−3a²b+3ab² = (a−b)³
Answered by jkamla60
29

Answer:

1 − 64a³ − 12a + 48a²

=(1)³ − (4a)³ − 3(1)²(4a) + 3(1)(4a)²

=(1 − 4a)²

=(1 − 4a)(1 − 4a)(1 − 4a)

Using the identity,a³−b³−3a²b+3ab² = (a−b)

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