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

Answer 1

To rewrite \( 40\sqrt[3]{3} \) as a product of an integer and another irrational cube root, we start by factoring \( 40 \):

\[
40 = 8 \times 5
\]

This allows us to express \( 40\sqrt[3]{3} \) as:

\[
40\sqrt[3]{3} = (8 \cdot 5)\sqrt[3]{3} = 8\sqrt[3]{3} \cdot 5
\]

Now using the property of cube roots, we can write:

\[
\sqrt[3]{40\sqrt[3]{3}} = \sqrt[3]{(8)(5)(\sqrt[3]{3})}= \sqrt[3]{8} \cdot \sqrt[3]{5} \cdot \sqrt[3]{\sqrt[3]{3}} = 2 \cdot \sqrt[3]{5} \cdot 3^{1/3}
\]

However, that may not seem very clear without simplifying properly.

To keep it as a more straightforward expression, we then want to express it in this format:

\[
x \cdot \sqrt[3]{y}
\]

We can approximate the form by collecting terms:

So the best option from your responses appears to be:

\[
2\sqrt[3]{10}
\]

This breakdown holds valid since \( 40\sqrt[3]{3} = 2 * (20\sqrt[3]{3}) = 2 * \sqrt[3]{20^3 \cdot 3} = 2\sqrt[3]{40} \).

So the closest matching response to the suitable format is:

**\( 2\sqrt[3]{10} \)**