Factorise: ${(a + 2 b - 3 c)}^{3} - a^{3} - 8 b^{3} + 27 c^{3}$ .

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Solution

Taking $x = a, y = 2 b, z = - 3 c$ , this is in the form
${(x + y + z)}^{3} - x^{3} - y^{3} - z^{3}$
We use factorisation
${(x + y + z)}^{3} - x^{3} - y^{3} - z^{3} = 3 (x + y) (y + z) (z + x)$
Putting back the values of $x, y, z$ , we obtain
${(a + 2 b - 3 c)}^{3} - a^{3} - 8 b^{3} + 27 c^{3} = 3 (a + 2 b) (2 b - 3 c) (a - 3 c)$

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