Question

What is the irrational cube root 40−−√3
rewritten as a product of an integer and another irrational cube root?(1 point)
Responses

25–√3
2 root index 3 start root 5 end root

210−−√3
2 root index 3 start root 10 end root

85–√3
8 root index 3 start root 5 end root

−25–√3
negative 2 root index 3 start root 5 end root

Answer 1

To rewrite the expression \( 40 - \sqrt{3} \) as a product of an integer and another irrational cube root, we need to rewrite \( 40 \) in a way that showcases a cube root component.

First, observe that \( 40 \) can be broken down as follows:
\[
40 = 8 \times 5
\]
Since \( 8 \) is a perfect cube, we can express this as:
\[
40 - \sqrt{3} = 8 \times 5 - \sqrt{3}
\]

We recognize that \( 8 = 2^3 \), hence,
\[
40 - \sqrt{3} = 2^3 \cdot 5 - \sqrt{3}
\]

Now, we want to isolate \(\sqrt[3]{5}\). The best way to rearrange this to express it as a product is to take out \( 2 \) or \( 2^3 \), which leads us to:

So the expression remains as:
\[
40 - \sqrt{3} = 2 \cdot \left( 4 \sqrt[3]{5} - \frac{\sqrt{3}}{2} \right)
\]
This type of manipulation may not yield a clean result; the goal was to showcase an integer times a cube root effectively.

The options provided are:
1. \(25 - \sqrt{3}\)
2. \(2 \sqrt[3]{10}\)
3. \(8 \sqrt[3]{5}\)
4. \(-2 \sqrt[3]{5}\)

The best candidate here matching the original form and correct factoring is:

**\( 8 \sqrt[3]{5} \)**, which simplifies the representation while clearly revealing irrational cube root involvement. Thus the valid response is:

**\( 8 \sqrt[3]{5} \)**.