Factorise:
${(2 x + y - z)}^{3} - 8 x^{3} - y^{3} + z^{3}$

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Solution

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

Using the above identity taking $a = 2 x$ , $b = y$ and $c = - z$ , the equation $(2 x + y - z)^{3} - 8 x^{3} - y^{3} + z^{3}$ can be factorised as follows:

$(2 x + y - z)^{3} - 8 x^{3} - y^{3} + z^{3} = 3 (2 x + y) (y - z) (- z + 2 x) = 3 (2 x + y) (y - z) (2 x - z)$
Hence, $(2 x + y - z)^{3} - 8 x^{3} - y^{3} + z^{3} = 3 (2 x + y) (y - z) (2 x - z)$

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