the simplified form of ∛54 is 3∛2 why is there a coefficient of 3 outside the cube root

1 answer

To simplify ∛54 (the cube root of 54), we can factor 54 to see if it has any perfect cubes.

The prime factorization of 54 is:

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

Since \(27\) is a perfect cube (\(3^3\)), we can simplify the cube root as follows:

\[ \sqrt[3]{54} = \sqrt[3]{27 \times 2} = \sqrt[3]{27} \times \sqrt[3]{2} \]

Now, calculate the cube root of \(27\):

\[ \sqrt[3]{27} = 3 \]

So we have:

\[ \sqrt[3]{54} = 3 \times \sqrt[3]{2} \]

Thus, the simplified form of \(\sqrt[3]{54}\) is:

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

The coefficient of \(3\) outside the cube root comes from the fact that \(27\) is equal to \(3^3\), which allows us to take the cube root of \(27\) and obtain the \(3\) as a coefficient.