Question

simplify the following expression.
$4sqrt3{54}+sqrt3{128}$
a $20sqrt3{2}$
b $16sqrt3{2}$
c $4sqrt3{182}$
d $36sqrt3{6}+8sqrt3{2}$

Explanation:

Step1: Simplify \(4\sqrt[3]{54}\)

First, factor 54: \(54 = 27\times2 = 3^3\times2\). So, \(\sqrt[3]{54}=\sqrt[3]{3^3\times2}=3\sqrt[3]{2}\). Then, \(4\sqrt[3]{54}=4\times3\sqrt[3]{2}=12\sqrt[3]{2}\).

Step2: Simplify \(\sqrt[3]{128}\)

Factor 128: \(128 = 64\times2 = 4^3\times2\). So, \(\sqrt[3]{128}=\sqrt[3]{4^3\times2}=4\sqrt[3]{2}\).

Step3: Add the two simplified terms

Now, add \(12\sqrt[3]{2}\) and \(4\sqrt[3]{2}\): \(12\sqrt[3]{2}+4\sqrt[3]{2}=(12 + 4)\sqrt[3]{2}=16\sqrt[3]{2}\).

Answer:

B. \(16\sqrt[3]{2}\)