Question

write the following product in simplest form.
\sqrt3{16} \cdot \sqrt3{-48}
\sqrt3{16} \cdot \sqrt3{-48} = \square
(simplify your answer. type an exact answer, using radicals as

Explanation:

Step1: Combine the cube roots

$\sqrt[3]{16} \cdot \sqrt[3]{-48} = \sqrt[3]{16 \times (-48)}$

Step2: Calculate the product inside

$\sqrt[3]{-768}$

Step3: Factor out perfect cubes

$\sqrt[3]{-64 \times 12} = \sqrt[3]{(-4)^3 \times 12}$

Step4: Simplify the cube root

$\sqrt[3]{(-4)^3} \cdot \sqrt[3]{12} = -4\sqrt[3]{12}$

Answer:

$-4\sqrt[3]{12}$