Eight Simplify Radical Expressions Problems Worksheet for 8th ... - Free Printable

Let's solve each of the simplification problems step by step. The goal is to simplify each expression using exponent rules.

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1) $ 4^{1/3} \cdot x^{1/3} \cdot \sqrt[3]{2x^2 \cdot y^{-1}} $

We rewrite all terms with exponents:

- $ 4^{1/3} = (2^2)^{1/3} = 2^{2/3} $
- $ x^{1/3} $
- $ \sqrt[3]{2x^2 y^{-1}} = (2x^2 y^{-1})^{1/3} = 2^{1/3} \cdot x^{2/3} \cdot y^{-1/3} $

Now multiply everything together:

$$
2^{2/3} \cdot x^{1/3} \cdot 2^{1/3} \cdot x^{2/3} \cdot y^{-1/3}
$$

Combine like terms:

- $ 2^{2/3 + 1/3} = 2^1 = 2 $
- $ x^{1/3 + 2/3} = x^1 = x $
- $ y^{-1/3} $

So the simplified form is:

$$
\boxed{2x \cdot y^{-1/3}} \quad \text{or} \quad \boxed{\frac{2x}{\sqrt[3]{y}}}
$$

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2) $ a^{-1} \cdot d^{1/2} \cdot a^{-1/2} \cdot b^{-1} $

Group like bases:

- $ a^{-1} \cdot a^{-1/2} = a^{-1 - 1/2} = a^{-3/2} $
- $ d^{1/2} $
- $ b^{-1} $

So we have:

$$
a^{-3/2} \cdot d^{1/2} \cdot b^{-1}
$$

Rewriting with positive exponents:

$$
\boxed{\frac{\sqrt{d}}{a^{3/2} \cdot b}} \quad \text{or} \quad \boxed{\frac{\sqrt{d}}{b \cdot a^{3/2}}}
$$

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3) $ 5m^{-1/2} \cdot x^{-1} \cdot \sqrt[3]{40m^{-2} \cdot x^{-1} \cdot y^{-1}} $

First, simplify the cube root:

- $ \sqrt[3]{40m^{-2}x^{-1}y^{-1}} = (40)^{1/3} \cdot m^{-2/3} \cdot x^{-1/3} \cdot y^{-1/3} $

Now write $ 40 = 8 \cdot 5 = 2^3 \cdot 5 $, so:

- $ 40^{1/3} = (2^3 \cdot 5)^{1/3} = 2 \cdot 5^{1/3} $

So cube root becomes:
$$
2 \cdot 5^{1/3} \cdot m^{-2/3} \cdot x^{-1/3} \cdot y^{-1/3}
$$

Now multiply by the outside terms:

- $ 5m^{-1/2} \cdot x^{-1} \cdot [2 \cdot 5^{1/3} \cdot m^{-2/3} \cdot x^{-1/3} \cdot y^{-1/3}] $

Multiply coefficients and combine exponents:

- Coefficients: $ 5 \cdot 2 \cdot 5^{1/3} = 10 \cdot 5^{1/3} $
- $ m^{-1/2 - 2/3} = m^{-3/6 - 4/6} = m^{-7/6} $
- $ x^{-1 - 1/3} = x^{-4/3} $
- $ y^{-1/3} $

So overall:

$$
10 \cdot 5^{1/3} \cdot m^{-7/6} \cdot x^{-4/3} \cdot y^{-1/3}
$$

We can write $ 10 \cdot 5^{1/3} = 10 \cdot \sqrt[3]{5} $, and express negative exponents as denominators:

$$
\boxed{\frac{10 \cdot \sqrt[3]{5}}{m^{7/6} \cdot x^{4/3} \cdot y^{1/3}}}
$$

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4) $ x^{1/2} \cdot y^{-1} \cdot \sqrt[3]{40x^{-2} \cdot y^{-1} \cdot z^{-1}} $

Again, simplify the cube root:

- $ \sqrt[3]{40x^{-2}y^{-1}z^{-1}} = 40^{1/3} \cdot x^{-2/3} \cdot y^{-1/3} \cdot z^{-1/3} $

As before, $ 40^{1/3} = 2 \cdot 5^{1/3} $

Now multiply by outside terms:

- $ x^{1/2} \cdot y^{-1} \cdot 2 \cdot 5^{1/3} \cdot x^{-2/3} \cdot y^{-1/3} \cdot z^{-1/3} $

Combine:

- Coefficient: $ 2 \cdot 5^{1/3} $
- $ x^{1/2 - 2/3} = x^{3/6 - 4/6} = x^{-1/6} $
- $ y^{-1 - 1/3} = y^{-4/3} $
- $ z^{-1/3} $

So:

$$
2 \cdot 5^{1/3} \cdot x^{-1/6} \cdot y^{-4/3} \cdot z^{-1/3}
$$

In fractional form:

$$
\boxed{\frac{2 \cdot \sqrt[3]{5}}{x^{1/6} \cdot y^{4/3} \cdot z^{1/3}}}
$$

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5) $ 2 \cdot m^{-1} \cdot \sqrt{m^{-1}} $

First, $ \sqrt{m^{-1}} = m^{-1/2} $

So:

$$
2 \cdot m^{-1} \cdot m^{-1/2} = 2 \cdot m^{-1 - 1/2} = 2 \cdot m^{-3/2}
$$

Or:

$$
\boxed{\frac{2}{m^{3/2}}} \quad \text{or} \quad \boxed{\frac{2}{\sqrt{m^3}}}
$$

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6) $ \sqrt[3]{x^{-2} \cdot y^{-1}} $

This is already a single radical:

$$
(x^{-2} y^{-1})^{1/3} = x^{-2/3} \cdot y^{-1/3}
$$

So:

$$
\boxed{\frac{1}{x^{2/3} \cdot y^{1/3}}} \quad \text{or} \quad \boxed{\frac{1}{\sqrt[3]{x^2 y}}}
$$

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✔ Final Answers:

1) $ \boxed{\frac{2x}{\sqrt[3]{y}}} $

2) $ \boxed{\frac{\sqrt{d}}{b \cdot a^{3/2}}} $

3) $ \boxed{\frac{10 \cdot \sqrt[3]{5}}{m^{7/6} \cdot x^{4/3} \cdot y^{1/3}}} $

4) $ \boxed{\frac{2 \cdot \sqrt[3]{5}}{x^{1/6} \cdot y^{4/3} \cdot z^{1/3}}} $

5) $ \boxed{\frac{2}{m^{3/2}}} $

6) $ \boxed{\frac{1}{\sqrt[3]{x^2 y}}} $

Let me know if you'd like these rewritten in other forms (e.g., with radicals instead of exponents)!