Question

what is the irrational cube root ^3√40 rewritten as a product of an integer and another irrational cube root?

8^3√5
2^3√5
2^3√10
-2^3√5

Answer 1

To rewrite \( \sqrt[3]{40} \) as a product of an integer and another irrational cube root, we can start by factoring \( 40 \):

\[
40 = 8 \times 5
\]

Now, we can express the cube root:

\[
\sqrt[3]{40} = \sqrt[3]{8 \times 5} = \sqrt[3]{8} \times \sqrt[3]{5}
\]

Since \( \sqrt[3]{8} = 2 \):

\[
\sqrt[3]{40} = 2 \times \sqrt[3]{5}
\]

So, the expression for \( \sqrt[3]{40} \) rewritten as a product of an integer and another irrational cube root is:

\[
2 \sqrt[3]{5}
\]

Among the options you provided, the correct answer is:

\[
\boxed{2 \sqrt[3]{5}}
\]