Chapter 5, Problem 2CR

To determine

To give the letter of the correct answer of the given function: $\frac{{\left(a^{2}b{c}^3\right)}^3}{{\left(-a{b}^3{c}^2\right)}^2}$

Answer to Problem 2CR

The correct answer is d.

Explanation of Solution

Given that: : $\frac{{\left(a^{2}b{c}^3\right)}^3}{{\left(-a{b}^3{c}^2\right)}^2}$

Apply law of Exponent: ${\left(xyz\right)}^n=x^n{y}^n{z}^n$

  $\frac{{\left(a^{2}b{c}^3\right)}^3}{{\left(-a{b}^3{c}^2\right)}^2}=\frac{{\left(a^2\right)}^3{\left(b\right)}^3{\left(c^3\right)}^3}{{\left(-a\right)}^2{\left(b^3\right)}^2{\left(c^2\right)}^2}$

Now, apply law of Exponents ${\left(a^n\right)}^m=a^{nm}$

  $\begin{array}{l}\frac{{\left(a^{2}b{c}^3\right)}^3}{{\left(-a{b}^3{c}^2\right)}^2}=\frac{{\left(a^2\right)}^3{\left(b\right)}^3{\left(c^3\right)}^3}{{\left(-a\right)}^2{\left(b^3\right)}^2{\left(c^2\right)}^2}\\ =\frac{\left(a^6\right)\left(b^3\right)\left(c^9\right)}{\left(a^2\right)\left(b^6\right)\left(c^4\right)}\end{array}$

Now, apply law of Exponents $\frac{a^n}{a^m}=a^{n-m}$ and simplify

  $\begin{array}{l}\Rightarrow \frac{\left(a^6\right)\left(b^3\right)\left(c^9\right)}{\left(a^2\right)\left(b^6\right)\left(c^4\right)}\\ =a^{\left(6-2\right)}\left(\frac{1}{b^{6-3}}\right)c^{(9-4)}\\ =\frac{a^4{c}^5}{b^3}\end{array}$

Hence, the correct answer is d.