Expand and simplify the following: \( \begin{array}{lll}\text { (a) }(x+4)^{2} & \text { (b) }(x-4)^{2} & \text { (c) }(2 x+4 y)^{2} \\ \text { (d) }(2 x-4 y)^{2} & \text { (c) }(-5 a-2 b)^{2} & \text { (f) }(a-3 b)^{3}\end{array} \)

Sure! Let's expand and simplify each expression step by step. (a) \((x + 4)^{2}\) Using the formula \((a + b)^{2} = a^{2} + 2ab + b^{2}\): \((x + 4)^{2} = x^{2} + 2(4)(x) + 4^{2} = x^{2} + 8x + 16\) (b) \((x - 4)^{2}\) Using \((a - b)^{2} = a^{2} - 2ab + b^{2}\): \((x - 4)^{2} = x^{2} - 2(4)(x) + 4^{2} = x^{2} - 8x + 16\) (c) \((2x + 4y)^{2}\) Using the expansion formula: \((2x + 4y)^{2} = (2x)^{2} + 2(2x)(4y) + (4y)^{2} = 4x^{2} + 16xy + 16y^{2}\) (d) \((2x - 4y)^{2}\) Using the formula for the difference: \((2x - 4y)^{2} = (2x)^{2} - 2(2x)(4y) + (4y)^{2} = 4x^{2} - 16xy + 16y^{2}\) (e) \((-5a - 2b)^{2}\) Using the square of a binomial: \((-5a - 2b)^{2} = (-5a)^{2} + 2(-5a)(-2b) + (-2b)^{2} = 25a^{2} + 20ab + 4b^{2}\) (f) \((a - 3b)^{3}\) Using the expansion for cubes: \((a - 3b)^{3} = a^{3} - 3(3b)a^{2} + 3(3b)^{2}a - (3b)^{3} = a^{3} - 9ab^{2} + 27b^{3}\) So, the expanded and simplified forms are: (a) \(x^{2} + 8x + 16\) (b) \(x^{2} - 8x + 16\) (c) \(4x^{2} + 16xy + 16y^{2}\) (d) \(4x^{2} - 16xy + 16y^{2}\) (e) \(25a^{2} + 20ab + 4b^{2}\) (f) \(a^{3} - 9ab^{2} + 27b^{3}\)