# Factorize the fol

(i) Given that: 1 β 64a3 β 12a + 48a2

= (1)3 + (β4a)3 + 3(1)2(β4a) + 3(1)(β4a)2

= (1 + (β4a))3

[β΅ a3 + b3 + 3a2b + 3ab2 = (a + b)3]

= (1 β 4a)3

=

= (2p)3 + (15)3 + 3(2p)2(15) + 3(2p) (15)2

= (2p + 15)3

[β΅ a3 + b3 + 3a2b + 3ab2 = (a + b) 3]

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