Question

$\frac{(4ab{)}^2}{-6a^2}÷\frac{(-a^{3}b{)}^2}{{2ab}^2}$ equals

• A. $\frac{{4b}^3}{{3a}^2}$
• B. $\frac{-16}{{3a}^3}$
• C. $\frac{{-16b}^2}{{3a}^5}$
• D. $\frac{{-8b}^2}{{3a}^5}$

Answer 1

The correct option is C $\frac{{-16b}^2}{{3a}^5}$

$\frac{(4ab{)}^2}{-6a^2}÷\frac{(-a^{3}b{)}^2}{{2ab}^2}$

$=\frac{16a^2{b}^2}{-6a^2}÷\frac{a^6{b}^2}{{2ab}^2}$

$=\frac{8b^2}{-3}÷\frac{a^5}{2}$

$=\frac{8b^2}{-3}\times \frac{2}{a^5}$

$=\frac{{-16b}^2}{{3a}^5}$