Question

simplify $6sqrt3{4} cdot 8sqrt3{28}$
\bigcirc $14sqrt3{32}$
\bigcirc $48sqrt3{112}$
\bigcirc $48sqrt3{14}$
\bigcirc $96sqrt3{14}$

Explanation:

Step1: Multiply the coefficients

First, multiply the coefficients (the numbers outside the cube roots) together. So, \(6\times8 = 48\).

Step2: Multiply the radicands

Next, multiply the radicands (the numbers inside the cube roots) together. That is, \(\sqrt[3]{4}\times\sqrt[3]{28}=\sqrt[3]{4\times28}\). Calculate \(4\times28 = 112\), so we have \(\sqrt[3]{112}\).

Step3: Combine the results

Now, combine the coefficient and the radicand. So, \(6\sqrt[3]{4}\cdot8\sqrt[3]{28}=48\sqrt[3]{112}\). But we can simplify \(\sqrt[3]{112}\) further. Factor \(112\) as \(8\times14\), and \(\sqrt[3]{8\times14}=\sqrt[3]{8}\times\sqrt[3]{14}=2\sqrt[3]{14}\). Then, \(48\times2\sqrt[3]{14}=96\sqrt[3]{14}\)? Wait, no, wait. Wait, let's check again. Wait, \(4\times28 = 112\), and \(112 = 8\times14\), so \(\sqrt[3]{112}=\sqrt[3]{8\times14}=2\sqrt[3]{14}\). Then, the original expression: \(6\times8 = 48\), and \(\sqrt[3]{4}\times\sqrt[3]{28}=\sqrt[3]{112}=\sqrt[3]{8\times14}=2\sqrt[3]{14}\). So then, \(48\times2\sqrt[3]{14}=96\sqrt[3]{14}\)? Wait, but let's check the steps again. Wait, maybe I made a mistake. Wait, the problem is \(6\sqrt[3]{4}\cdot8\sqrt[3]{28}\). So first, multiply the coefficients: \(6\times8 = 48\). Then multiply the radicands: \(\sqrt[3]{4}\times\sqrt[3]{28}=\sqrt[3]{4\times28}=\sqrt[3]{112}\). Now, simplify \(\sqrt[3]{112}\): \(112 = 8\times14\), so \(\sqrt[3]{8\times14}=\sqrt[3]{8}\times\sqrt[3]{14}=2\sqrt[3]{14}\). So now, multiply the coefficient \(48\) by \(2\) (from the cube root simplification): \(48\times2 = 96\), and then we have \(\sqrt[3]{14}\). So the simplified form is \(96\sqrt[3]{14}\). Wait, but let's check the options. The last option is \(96\sqrt[3]{14}\). Let's verify again.

First, \(6\sqrt[3]{4} \times 8\sqrt[3]{28}\):

• Coefficients: \(6 \times 8 = 48\)
• Radicands: \(\sqrt[3]{4} \times \sqrt[3]{28} = \sqrt[3]{4 \times 28} = \sqrt[3]{112}\)
• Now, factor \(112\) into prime factors: \(112 = 2^4 \times 7 = 2^3 \times 2 \times 7 = 8 \times 14\)
• So, \(\sqrt[3]{112} = \sqrt[3]{8 \times 14} = \sqrt[3]{8} \times \sqrt[3]{14} = 2\sqrt[3]{14}\)
• Now, multiply the coefficient \(48\) by \(2\) (from the cube root of 8): \(48 \times 2 = 96\)
• So, putting it together: \(96\sqrt[3]{14}\)

Yes, that's correct. So the correct answer is \(96\sqrt[3]{14}\), which is the last option.

Answer:

\(96\sqrt[3]{14}\) (corresponding to the option: \(96\sqrt[3]{14}\))