\( \frac { ( 3 b ) ^ { 2 } + 12 a b ^ { 5 } } { 3 b } = 6 b + \square a b \)

We start with \[ \frac{(3b)^2+12ab^5}{3b}. \] **Step 1. Expand the numerator** Calculate \((3b)^2\): \[ (3b)^2=9b^2. \] So the numerator becomes \[ 9b^2+12ab^5. \] **Step 2. Divide each term by \(3b\)** Write the expression as \[ \frac{9b^2}{3b}+\frac{12ab^5}{3b}. \] Simplify each term: - For the first term: \[ \frac{9b^2}{3b} = 3b. \] - For the second term: \[ \frac{12ab^5}{3b} = 4a\,b^{5-1} = 4ab^4. \] Thus, the simplified result is \[ 3b+4ab^4. \] **Step 3. Write \(4ab^4\) in the form “\(\square\, ab\)”** We wish to express the term \(4ab^4\) in the form \[ \square\,ab. \] Notice that \[ 4ab^4 = 4b^3\cdot ab, \] since \[ ab\cdot b^3 = ab^4. \] Therefore, we have \[ \frac{(3b)^2+12ab^5}{3b} = 3b+4ab^4 = 3b+ (4b^3)ab. \] Comparing with the form \[ 6b+\square\, ab, \] we see that if we want consistency the term multiplying \(ab\) must carry the factor \(4b^3\). (One may note that the constant term here is \(3b\) rather than \(6b\); thus, if the textbook expression is intended to have \(6b\) as the first term, then a correction might be needed. But as given, the simplification shows the missing factor in the \(ab\) term is \(4b^3\).) **Final Answer:** \[ \square = 4b