Question

The simplfied form of 54−−√3
is 32–√3
. Why is there a coefficient of 3 outside the cube root?(1 point)
Responses

There is a coefficient of 3 outside the cube root because 54 is divisible by 3.
There is a coefficient of 3 outside the cube root because 54 is divisible by 3.

There is a coefficient of 3 outside the cube root because three 3s divide into 54.
There is a coefficient of 3 outside the cube root because three 3s divide into 54.

There is a coefficient of 3 outside the cube root because two 3s divide into 54.
There is a coefficient of 3 outside the cube root because two 3s divide into 54.

There is a coefficient of 3 outside the cube root because 9 divides into 54.

Answers

Answered by GPT-4o mini
The correct explanation for the coefficient of 3 outside the cube root in the expression \( \sqrt[3]{54\sqrt{3}} \) can be identified by recognizing that \( 54 \) can be factored in such a way that a perfect cube can be extracted. The number \( 54 \) can be factored as \( 27 \times 2 \) (since \( 27 \) is \( 3^3 \)) or \( 9 \times 6 \).

To simplify, we can write:

\[
54 = 27 \times 2 = 3^3 \times 2
\]

Thus,

\[
\sqrt[3]{54\sqrt{3}} = \sqrt[3]{(27 \cdot 2)(\sqrt{3})} = \sqrt[3]{27} \cdot \sqrt[3]{2\sqrt{3}} = 3 \cdot \sqrt[3]{2\sqrt{3}}
\]

Notice in the simplifications, one factor of \( 3 \) comes from \( 27 \) (which simplifies to 3 outside the cube root).

So, the appropriate response to why there is a coefficient of 3 outside the cube root is:

**There is a coefficient of 3 outside the cube root because three 3s divide into 54.**

This is the most accurate statement, as it emphasizes how the factorization leads to extracting a cube root.

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