mohammedrazeenmn mohammedrazeenmn
  • 29-05-2021
  • Mathematics
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.−64a3+48a2−12a+1

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grxkdncc99
grxkdncc99 grxkdncc99
  • 29-05-2021
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)³
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