Question

Use proper identity to factorise 1+64a3

Answer 1

Answer:

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Answer 2

Answer:

$1 + 64a \: \ ^{3} \\ = 1 ^{3} + (4a) \:^{3} \\ = (1 + 4a) \: [\: (4a)^{2} ) - (4a \times 1) + (1) ^{2} ]\\ = (1 + 4a)(16a ^{2} - 4a + 1)$

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