Q. 254 - Intermediate Algebra

Question

In the following exercises, simplify.

$$\begin{gathered} a) \frac{\sqrt{810c^{-3}{d}^7}}{\sqrt{1000c{d}^{-1}}} \\ b) \frac{\sqrt[3]{24a^7{b}^{-1}}}{\sqrt[3]{-81a^{-2}{b}^2}} \end{gathered}$$

Step-by-Step Solution

Answer

Part(a) the simplified form is $\frac{9d^4}{10c^2}$.

Part(b) the simplified form is $-\frac{2a^3}{3b}$.

Step 1 Given information.

The given expressions are,

$$\begin{gathered} a) \frac{\sqrt{810c^{-3}{d}^7}}{\sqrt{1000c{d}^{-1}}} \\ b) \frac{\sqrt[3]{24a^7{b}^{-1}}}{\sqrt[3]{-81a^{-2}{b}^2}} \end{gathered}$$

Step 2. Solving part(a).

Now, to simplify the expression first we will rewrite the expression using the quotient property.

$\frac{\sqrt{810c^{-3}{d}^7}}{\sqrt{1000c{d}^{-1}}}=\sqrt{\frac{810c^{-3}{d}^7}{1000c{d}^{-1}}}$

Then removing the common factors.

$\sqrt{\frac{810c^{-3}{d}^7}{1000c{d}^{-1}}}=\sqrt{\frac{81d^8}{100c^4}}$

On simplifying we get,

$\sqrt{\frac{81d^8}{100c^4}}=\frac{9d^4}{10c^2}$

Hence, the simplified form is $\frac{9d^4}{10c^2}$.

Step 3. Solving part(b).

Now, to simplify the expression first we will rewrite the expression using the quotient property.

$\frac{\sqrt[3]{24a^7{b}^{-1}}}{\sqrt[3]{-81a^{-2}{b}^2}}=\sqrt[3]{\frac{24a^7{b}^{-1}}{-81a^{-2}{b}^2}}$

Then removing the common factors.

$\sqrt[3]{\frac{24a^7{b}^{-1}}{-81a^{-2}{b}^2}}=\sqrt[3]{\frac{8a^9}{-27b^3}}$

On simplifying we get,

$\sqrt[3]{\frac{8a^9}{-27b^3}}=-\frac{2a^3}{3b}$

Hence, the simplified form is $-\frac{2a^3}{3b}$.