Question

simplify the expression $4sqrt3{-9} cdot 7sqrt3{48}$

Explanation:

Step1: Multiply the coefficients and combine the cube roots

First, multiply the coefficients \(4\) and \(7\), and use the property of radicals \(\sqrt[3]{a}\cdot\sqrt[3]{b}=\sqrt[3]{ab}\) for the cube roots. So we have:
\(4\sqrt[3]{-9}\cdot7\sqrt[3]{48}=(4\times7)\cdot\sqrt[3]{-9\times48}\)
\( = 28\sqrt[3]{-432}\)

Step2: Simplify the cube root

Now, factor \(-432\) to find a perfect cube factor. We know that \(-432=-8\times54=-8\times27\times2\), and \(\sqrt[3]{-8}=-2\), \(\sqrt[3]{27} = 3\). So:
\(\sqrt[3]{-432}=\sqrt[3]{-8\times27\times2}=\sqrt[3]{-8}\times\sqrt[3]{27}\times\sqrt[3]{2}=-2\times3\times\sqrt[3]{2}=-6\sqrt[3]{2}\)

Step3: Multiply with the coefficient

Now multiply this result with the coefficient \(28\) from Step 1:
\(28\times(-6\sqrt[3]{2})=-168\sqrt[3]{2}\)

Answer:

\(-168\sqrt[3]{2}\)