Question
The simplfied form of root(54, 3) is 3 * root(2, 3) Why is there a coefficient of 3 outside the cube root? (1 point ) 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 9 divides 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 54 is divisible by 3
Answers
Answered by
GPT-4o mini
To simplify the cube root of 54, we can factor 54 into its prime factors:
54 can be factored as \( 54 = 2 \times 3^3 \).
When taking the cube root of 54, we can separate the factors:
\[
\sqrt[3]{54} = \sqrt[3]{2 \times 3^3} = \sqrt[3]{2} \times \sqrt[3]{3^3}
\]
The cube root of \(3^3\) is 3, so we have:
\[
\sqrt[3]{54} = 3 \sqrt[3]{2}
\]
Thus, the coefficient of 3 outside the cube root appears because \(3^3\) (which equals 27) can be factored out of 54. Hence, the correct statement to support this conclusion is:
"There is a coefficient of 3 outside the cube root because two 3s divide into 54."
54 can be factored as \( 54 = 2 \times 3^3 \).
When taking the cube root of 54, we can separate the factors:
\[
\sqrt[3]{54} = \sqrt[3]{2 \times 3^3} = \sqrt[3]{2} \times \sqrt[3]{3^3}
\]
The cube root of \(3^3\) is 3, so we have:
\[
\sqrt[3]{54} = 3 \sqrt[3]{2}
\]
Thus, the coefficient of 3 outside the cube root appears because \(3^3\) (which equals 27) can be factored out of 54. Hence, the correct statement to support this conclusion is:
"There is a coefficient of 3 outside the cube root because two 3s divide into 54."