Products of Irrational Square Roots Quick Check

Question

4 of 54 of 5 Items
Question
What is the irrational cube root ^3√-24
rewritten as a product of an integer and another irrational cube root?(1 point)
Responses

−2^3√4
negative 2 root index 3 start root 4 end root

−2^3√3
negative 2 root index 3 start root 3 end root

2^3√6
2 root index 3 start root 6 end root

2^3√3

Answer 1

To rewrite the irrational cube root \( \sqrt[3]{-24} \) as a product of an integer and another irrational cube root, we first note that:

\[ -24 = -1 \times 24 = -1 \times (8 \times 3) = -1 \times 8 \times 3 \]

Then, we can separate the cube root as follows:

\[ \sqrt[3]{-24} = \sqrt[3]{-1 \times 8 \times 3} = \sqrt[3]{-1} \times \sqrt[3]{8} \times \sqrt[3]{3} \]

We know that:

\[ \sqrt[3]{-1} = -1 \quad \text{and} \quad \sqrt[3]{8} = 2 \]

Thus, we can rewrite it as:

\[ \sqrt[3]{-24} = -1 \times 2 \times \sqrt[3]{3} = -2\sqrt[3]{3} \]

So, the correct choice is:

\[ \boxed{-2 \sqrt[3]{3}} \]