What is the irrational cube root 40−−√3

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:

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