Rewrite the irrational cube root 48−−√3

as a product of an integer and another irrational cube root.(1 point)
Responses

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

412−−√3
4 root index 3 start root 12 end root

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

43–√3

1 answer

To rewrite \( \sqrt[3]{48} \) as a product of an integer and another irrational cube root, we start by factoring \( 48 \):

\[ 48 = 16 \times 3 = 4^2 \times 3 \]

Now we can express the cube root:

\[ \sqrt[3]{48} = \sqrt[3]{16 \times 3} = \sqrt[3]{16} \cdot \sqrt[3]{3} \]

Next, we can write \( 16 \) as \( 4^2 \):

\[ \sqrt[3]{16} = \sqrt[3]{4^2} = 4^{2/3} \]

However, since we need to express it as a simple integer coefficient times another cube root, we can take out the perfect cube from \( 48 \):

Notably, we can rewrite \( 48 \) as follows:

\[ 48 = 8 \times 6 \]

where \( 8 \) is a perfect cube (\( 2^3 \)):

\[ \sqrt[3]{48} = \sqrt[3]{8 \times 6} = \sqrt[3]{8} \cdot \sqrt[3]{6} = 2 \cdot \sqrt[3]{6} \]

Thus, we can express \( \sqrt[3]{48} \) as:

\[ \sqrt[3]{48} = 2 \sqrt[3]{6} \]

Therefore, the correct response is:

2 root index 3 start root 6 end root