Factorise using suitable identity:
${8 a}^{3} - 27 b^{3} - 64 c^{3} - 72 a b c$

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Solution

$8 a^{3} - 27 b^{3} - 64 c^{3} - 72 a b c$

$= {(2 a)}^{3} + {(- 3 b)}^{3} + {(- 4 c)}^{3} - 3 \times (2 a) (- 3 b) (- 4 c)$

Using the identity $x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$

where $x = 2 a, y = - 3 b, z = - 4 c$

$= (2 a - 3 b - 4 c) ({4 a}^{2} + {9 b}^{2} + {16 c}^{2} + 6 a b - 12 b c + 8 a c)$

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