Question

What is the irrational cube root −24−−−−√3
rewritten as a product of an integer and another irrational cube root?(1 point)
Responses

23–√3
2 root index 3 start root 3 end root

−24–√3
negative 2 root index 3 start root 4 end root

26–√3
2 root index 3 start root 6 end root

−23–√3

Answer 1

To rewrite the irrational cube root \(-24\sqrt{3}\) as a product of an integer and another irrational cube root, we can factor \(-24\) as follows:

\[
-24 = -8 \times 3
\]

Next, we can express \(-24\sqrt{3}\) using this factorization:

\[
-24\sqrt{3} = (-8)(\sqrt{3}) = -8\sqrt{3}
\]

Now, since \( -8 \) can be rewritten as \( -2^3 \), we can express \(-8\) in terms of its cube root:

\[
\sqrt[3]{-8} = -2
\]

Thus:

\[
-24\sqrt{3} = (-2)^3 \times \sqrt{3} = -2 \cdot \sqrt[3]{3 \cdot 8} = -2 \sqrt[3]{24}
\]

However, to match the options provided, we realize we want to express \(-24\sqrt{3}\) in the simplest form related to the cube root:

Factoring out the cube root, we can say:

\[
-24\sqrt{3} = -2 \cdot \sqrt[3]{12 \cdot 3} = -2 \cdot \sqrt[3]{36}
\]

Thus, the most suitable option corresponding to our simplified form is:

**−2 \sqrt[3]{4}**

(However, the original direct terms might ensure changes based on different constraints; options can't be fully validated as they require examining structured format.)