Simplify each of the following expressions. Assume that \( b \neq 0 \). Verify, \( \begin{array}{ll}\text { a. } m^{4}+(2 m)^{4} & \text { d. } \frac{a b^{3}}{b}+(a b)^{2}-a b^{2}\end{array} \)

Let's simplify the given expressions step by step: a. For \( m^{4} + (2m)^{4} \): - First, simplify \( (2m)^{4} \): \[ (2m)^{4} = 2^{4} \cdot m^{4} = 16m^{4} \] - Now substitute this back into the expression: \[ m^{4} + 16m^{4} = 17m^{4} \] So the simplified expression for \( a \) is \( 17m^{4} \). d. For \( \frac{ab^{3}}{b} + (ab)^{2} - ab^{2} \): - Start with \( \frac{ab^{3}}{b} \): \[ \frac{ab^{3}}{b} = ab^{2} \quad \text{(since } b \neq 0\text{)} \] - Now, calculate \( (ab)^{2} \): \[ (ab)^{2} = a^{2}b^{2} \] - Substitute these into the expression: \[ ab^{2} + a^{2}b^{2} - ab^{2} \] - Combine like terms: \[ (ab^{2} - ab^{2}) + a^{2}b^{2} = 0 + a^{2}b^{2} = a^{2}b^{2} \] So the simplified expression for \( d \) is \( a^{2}b^{2} \). Final Answers: a. \( 17m^{4} \) d. \( a^{2}b^{2} \)