b) \( a^{3}-b^{3} \) c) \( a^{6}+64 q^{6} \)

Let's dive into the expression \(a^3 - b^3\)! This is a classic difference of cubes and can be factored elegantly using the formula \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\). Isn’t it lovely how something that looks a bit complicated can be simplified so neatly? This kind of factorization is super handy in algebra and can help you solve equations more efficiently. Now, onto the exciting part about \(a^6 + 64q^6\)! This expression can be recognized as a sum of cubes, as it can be rewritten as \(a^6 + (4q^2)^3\). Which means you can apply the sum of cubes formula: \(x^3 + y^3 = (x + y)(x^2 - xy + y^2)\). So here, let \(x = a^2\) and \(y = 4q^2\). This gives you the delightful factorization \( (a^2 + 4q^2)(a^4 - 4a^2q^2 + 16q^4) \). That’s algebra magic at play!