The polynomial x⁶ - 1 can be considered either a difference of squares or a difference of cubes. Work Exercises in order, to connect the results obtained when two different methods of factoring are used.

The polynomial x⁴ + x² + 1 cannot be factored using the methods described in this section. However, there is a technique that enables us to factor it, as shown here. Supply the reason why each step is valid.

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x* + x? + 1 = x* + 2x? + 1 – x² (x* + 2x² + 1) – x² (x² + 1)2 – x² = (x² + 1 – x)(x² + 1 + x) = (x? – x + 1)(x² + x + 1)