Factorise the following
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Problem 1.4 Factor (a+2b−3c)3+(b+2c−3a)3+(c+2a−3b)3.
Solution Observe that (a+2b−3c)+(b+2c−3a)+(c+2a−3b)=0. Because x+y+z=0 implies x 3+y 3+z 3=3xyz, we obtain
$$\begin{gathered} {\left( {a + 2b - 3c} \right)^3} + {\left( {b + 2c - 3a} \right)^3} + {\left( {c + 2a - 3b} \right)^3} \hfill \\ \quad = 3\left( {a + 2b - 3c} \right)\left( {b + 2c - 3a} \right)\left( {c + 2a - 3b} \right). \hfill \\ \end{gathered}$$
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