Question

factorise 1+64x power 3​

Answer 1

$1 + 64x {}^{3} \\ = 1 {}^{3} + (4x) {}^{3} \\ = (1 + 4x)(1 {}^{2} - 1 \times 4x + (4x) {}^{2} ) \\ = (1 + 4x)(1 - 4x + 16x {}^{2} )$

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