Factorise the following:
b(c - d)2 + a(d - c) + 3c - 3d
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b(c - d)2 + a(d - c) + 3c - 3d = b(c - d)2 + a(-1)(c - d) + 3(c - d)
= (c - d)[b(c - d) - a + 3]
= (c - d)(bc - bd - a + 3).
Hence, b(c - d)2 + a(d - c) + 3c - 3d = (c - d)(bc - bd - a + 3).
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