Factorise: $8 y^{3} - 1$

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Solution

We know the identity $a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b)$

Using the above identity, the equation $8 y^{3} - 1$ can be factorised as follows:

$8 y^{3} - 1 = (2 y)^{3} - (1)^{3} = (2 y - 1) [(2 y)^{2} + 1^{2} + (2 y \times 1)] = (2 y - 1) (4 y^{2} + 1 + 2 y)$

Hence, $8 y^{3} - 1 = (2 y - 1) (4 y^{2} + 1 + 2 y)$

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