Question

simplify the following expression.
$4\sqrt3{432} - 5\sqrt3{128}$
a $2\sqrt3{2}$
b $4\sqrt3{2}$
c $\sqrt3{2}$
d $48\sqrt3{3} - 40\sqrt3{2}$

Explanation:

Step1: Simplify $\sqrt[3]{432}$

Factor 432: $432 = 8\times 54 = 8\times 6\times 9 = 8\times 2\times 3\times 9 = 8\times 2\times 3\times 3\times 3 = 8\times 2\times 3^3$. So $\sqrt[3]{432}=\sqrt[3]{8\times 2\times 3^3}=\sqrt[3]{8}\times\sqrt[3]{3^3}\times\sqrt[3]{2}=2\times 3\times\sqrt[3]{2}=6\sqrt[3]{2}$.

Step2: Simplify $\sqrt[3]{128}$

Factor 128: $128 = 64\times 2 = 4^3\times 2$. So $\sqrt[3]{128}=\sqrt[3]{4^3\times 2}=4\sqrt[3]{2}$.

Step3: Substitute back into the original expression

The original expression is $4\sqrt[3]{432}-5\sqrt[3]{128}$. Substitute the simplified cube roots: $4\times 6\sqrt[3]{2}-5\times 4\sqrt[3]{2}=24\sqrt[3]{2}-20\sqrt[3]{2}$.

Step4: Combine like terms

$24\sqrt[3]{2}-20\sqrt[3]{2}=(24 - 20)\sqrt[3]{2}=4\sqrt[3]{2}$.

Answer:

B. $4\sqrt[3]{2}$