1. **State the problem:** Factorize the expression $$a^4 + 64b^4$$. 2. **Recall the formula:** This is a sum of two fourth powers. We can use the sum of squares and difference of squares formulas to factor it. 3. **Rewrite the expression:** Note that $$64b^4 = (2b)^4$$, so the expression is $$a^4 + (2b)^4$$. 4. **Use the sum of squares factorization:** $$a^4 + (2b)^4 = (a^2)^2 + (2b)^4$$ can be seen as $$x^2 + y^2$$ where $$x = a^2$$ and $$y = (2b)^2 = 4b^2$$. 5. **Apply the Sophie Germain identity:** For $$x^4 + 4y^4$$, the factorization is: $$x^4 + 4y^4 = (x^2 - 2xy + 2y^2)(x^2 + 2xy + 2y^2)$$. 6. **Match terms:** Here, $$a^4 + 64b^4 = a^4 + 4(2b)^4$$, so $$x = a$$ and $$y = 2b$$. 7. **Write the factorization:** $$a^4 + 64b^4 = (a^2 - 2a(2b) + 2(2b)^2)(a^2 + 2a(2b) + 2(2b)^2)$$ 8. **Simplify inside the factors:** - First factor: $$a^2 - 4ab + 8b^2$$ - Second factor: $$a^2 + 4ab + 8b^2$$ 9. **Final answer:** $$a^4 + 64b^4 = (a^2 - 4ab + 8b^2)(a^2 + 4ab + 8b^2)$$ This factorization cannot be simplified further over the real numbers.