Answer
(a) $$A^4-B^4 = (A-B)(A+B)(A^2+b^2)$$
$$A^6-B^6 = (A-B)(A^2+AB+B^2)(A+B)(A^2-AB+B^2)$$
(b) $$12^4-7^4=5\times19\times193=95\times193=18,335$$
$$12^6-7^6=5\times277\times19\times109=1385\times2071=2,868,335$$
(c) $$18,335 = 5\times19\times193$$
$$2,868,335 = 5\times277\times19\times109$$
Work Step by Step
(a) To easily factorize the expression we will use Special Factoring Formulas (p.30). In this case Difference of Squares.
$$A^4-B^4 =(A^2-B^2)(A^2+B^2)= (A-B)(A+B)(A^2+b^2)$$
This time we will use Difference of Squares at first and then Sum & Difference of Cubes.
$$A^6-B^6 =(A^3-B^3)(A^3+B^3)= (A-B)(A^2+AB+B^2)(A+B)(A^2-AB+B^2)$$
(b) Here we will use exactly the same formulas we used in (a), but instead of variables we have integers now. In first case Difference of Squares 2 times and then simply calculate
$$12^4-7^4=(12^2-7^2)(12^2+7^2) = (12-7)(12+7)(144+49) = 5\times 19\times193 =95\times193=18,335$$
We will use Difference of Squares and then Sum & Difference of Cubes
$$12^6-7^6=(12^3-7^3)(12^3+7^3) = (12-7) (144+12\times7+49) (12+7) (144-12\times7+49) =5\times(144+84+49) \times19\times(144-84+49) = 5\times277\times19\times109=1385\times2071=2,868,335$$
(c) While verifying expressions in (b) we came up with answer for (c). We got them factorized by prime numbers:
$$18,335 = 5\times19\times193$$
$$2,868,335 = 5\times277\times19\times109$$
