Question

Simplify the expression.

a) ${\left({\mathbf{a}}^2\right)}^{-3}{\left({\mathbf{a}}^3\mathbf{b}\right)}^2{\left({\mathbf{b}}^3\right)}^4$

b) ${\left(\mathbf{3}{\mathbf{xy}}^2\right)}^3{\left(\frac{2}{3}{\mathbf{x}}^{-1}\mathbf{y}\right)}^2$

Short answer

Answer is${\left(a^2\right)}^{-3}{\left(a^{3}b\right)}^2{\left(b^3\right)}^4=b^{14}$

Answer iswidth="191">3xy2323x−1y2=12x1y8

Step-by-step solution

Part a. Step 1. Identities of powers.

$\begin{array}{l}{\left(x^a\right)}^b=x^{ab}\\ x^a\times x^b=x^{a+b}\end{array}$

Part a. Step 2. Re write the expression after applying the identities.

${\left(a^2\right)}^{-3}{\left(a^{3}b\right)}^2{\left(b^3\right)}^4=\left(a^{-6}\right)\left(a^6{b}^2\right)\left(b^{12}\right)$

Part a. Step 3. Simplify the expression by solving the powers.

$$\begin{gathered} \left(a^{-6}\right)\left(a^6{b}^2\right)\left(b^{12}\right)=\left(a^{-6}{a}^6\right)\left(b^2{b}^{12}\right) \\ =\left(a^{-6+6}\right)\left(b^{2+12}\right) \\ =a^0{b}^{14} \\ =b^{14} \end{gathered}$$

Part b. Step 1. Identities of powers.

$\begin{array}{l}{\left(x^a\right)}^b=x^{ab}\\ x^a\times x^b=x^{a+b}\end{array}$

Part b. Step 2. Re write the expression after applying the identities.

${\left(3x{y}^2\right)}^3{\left(\frac{2}{3}{x}^{-1}y\right)}^2=27x^3{y}^6\left(\frac{4}{9}{x}^{-2}{y}^2\right)$

Part b. Step 3. Simplify the expression by solving the powers.

$$\begin{gathered} 27x^3{y}^6\left(\frac{4}{9}{x}^{-2}{y}^2\right)=\left(27\times \frac{4}{9}\right)\left(x^3{x}^{-2}\right)\left(y^6{y}^2\right) \\ =12\left(x^{3+\left(-2\right)}\right)\left(y^{6+2}\right) \\ =12x^1{y}^8 \end{gathered}$$