Question

9 simplify.
\frac{\sqrt3{270x^{20}}}{\sqrt3{5x}}
a \\ 3x^{6} \sqrt3{2x}
b \\ \sqrt3{135x^{19}}
c \\ 2x \sqrt3{3x^{6}}
d \\ 3x^{6} \sqrt{135x}

Explanation:

Step1: Use Quotient Rule for Radicals

The quotient rule for cube roots states that $\frac{\sqrt[3]{a}}{\sqrt[3]{b}} = \sqrt[3]{\frac{a}{b}}$. So we can combine the two cube roots into one: $\sqrt[3]{\frac{270x^{20}}{5x}}$.

Step2: Simplify the Fraction Inside the Radical

Simplify the coefficient and the variable separately. For the coefficient: $\frac{270}{5} = 54$. For the variable: using the rule $\frac{x^m}{x^n}=x^{m - n}$, we have $x^{20-1}=x^{19}$. So now we have $\sqrt[3]{54x^{19}}$.

Step3: Factor the Radicand into Perfect Cubes and Remainders

Factor 54: $54 = 27\times2$, and $27 = 3^3$. Factor $x^{19}$: $x^{19}=x^{18}\times x=(x^6)^3\times x$. So we can rewrite the radicand as $\sqrt[3]{3^3\times2\times(x^6)^3\times x}$.

Step4: Apply the Cube Root Property

The cube root of a product is the product of the cube roots: $\sqrt[3]{3^3\times(x^6)^3\times2x}=\sqrt[3]{3^3}\times\sqrt[3]{(x^6)^3}\times\sqrt[3]{2x}$.

Step5: Simplify Each Cube Root

$\sqrt[3]{3^3}=3$, $\sqrt[3]{(x^6)^3}=x^6$, and the last term remains $\sqrt[3]{2x}$. Multiplying these together gives $3x^6\sqrt[3]{2x}$.

Answer:

A. $3x^{6}\sqrt[3]{2x}$