Question

Simplify:$\left(a\right)\sqrt{225m^2{n}^2}\left(b\right)\sqrt{169x^{10}{y}^{14}}\left(c\right)\sqrt[3]{27w^{36}{z}^{15}}$

Short answer

The solutions of (a), (b) and (c) are $15\left|mn\right|,13\left|x^5{y}^7\right|and3w^{12}{z}^5$ respectively.

Step-by-step solution

Part (a) Step 1. Given information

The root: $\sqrt{225m^2{n}^2}$

Part (a) Step 2. Calculation

We have,

$$\begin{gathered} \sqrt{225m^2{n}^2} \\ =\sqrt{{\left(15mn\right)}^2}\sin ce{\left(15mn\right)}^2=225m^2{n}^2 \\ =15\left|mn\right| \end{gathered}$$

Therefore, the solution is$15\left|mn\right|$

Part (b) Step 1. Given information

The root:$\sqrt{169x^{10}{y}^{14}}$

Part (b) Step 2. Calculation

We have,

$$\begin{gathered} \sqrt{169x^{10}{y}^{14}} \\ =\sqrt{{\left(13x^5{y}^7\right)}^2}\sin ce{\left(13x^5{y}^7\right)}^2=169x^{10}{y}^{14} \\ =13\left|x^5{y}^7\right| \end{gathered}$$

Therefore, the solution is$13\left|x^5{y}^7\right|$

Part (c) Step 1. Given information

The root:$\sqrt[3]{27w^{36}{z}^{15}}$

Part (c) Step 2. Calculation

We have,

$$\begin{gathered} \sqrt[3]{27w^{36}{z}^{15}} \\ =\sqrt[3]{{\left(3w^{12}{z}^5\right)}^3}\sin ce{\left(3w^{12}{z}^5\right)}^3=27w^{36}{z}^{15} \\ =3w^{12}{z}^5 \end{gathered}$$

Therefore, the solution is$3w^{12}{z}^5$