The binomial x⁶ - y⁶ may be considered as either a difference of squares or a difference of cubes.

If we have a choice of factoring as a difference of squares or a difference of cubes, how should we start to more easily obtain the completely factored form of the polynomial? Base the answer on the results in Exercises 75–79.

Data from in Exercises 75

Factor x⁶ - y⁶ by first factoring as a difference of squares. Then factor further by considering one of the factors as a sum of cubes and the other factor as a difference of cubes.

Data from in Exercise 76

Based on the answer in Exercise 75, fill in the blank with the correct factors so that x⁶ - y⁶ is factored completely.

Data from in Exercise 77

Factor x⁶ - y⁶ by first factoring as a difference of cubes. Then factor further by considering one of the factors as a difference of squares.

Data from in Exercise 78

Based on the answer in Exercise 77, fill in the blank with the correct factor so that x⁶ - y⁶ is factored.

Data from in Exercise 79

Notice that the factor written in the blank in Exercise 78 is a fourth-degree polynomial, while the two factors written in the blank in Exercise 76 are both second-degree polynomials. What must be true about the product of the two factors written in the blank in Exercise 76? Verify this.

x6y6 = (x - y) (x + y).