Question

question
rewrite the following expression in simplest radical form given ( x > 0 ).
( sqrt4{16x} cdot sqrt3{216x^2} )

Explanation:

Step1: Simplify the fourth root

Simplify $\sqrt[4]{16x}$. We know that $16 = 2^4$, so $\sqrt[4]{16x}=\sqrt[4]{2^4\cdot x}=2\sqrt[4]{x}$.

Step2: Simplify the cube root

Simplify $\sqrt[3]{216x^2}$. We know that $216 = 6^3$, so $\sqrt[3]{216x^2}=\sqrt[3]{6^3\cdot x^2}=6\sqrt[3]{x^2}$.

Step3: Multiply the simplified radicals

Multiply the results from Step 1 and Step 2: $(2\sqrt[4]{x})\cdot(6\sqrt[3]{x^2}) = 12\sqrt[4]{x}\cdot\sqrt[3]{x^2}$.
To combine the radicals, we use the property $a^m\cdot a^n=a^{m + n}$ and rewrite the radicals with a common exponent. The least common multiple of 4 and 3 is 12. So, $\sqrt[4]{x}=x^{\frac{1}{4}}=x^{\frac{3}{12}}=\sqrt[12]{x^3}$ and $\sqrt[3]{x^2}=x^{\frac{2}{3}}=x^{\frac{8}{12}}=\sqrt[12]{x^8}$.
Then, $\sqrt[12]{x^3}\cdot\sqrt[12]{x^8}=\sqrt[12]{x^{3 + 8}}=\sqrt[12]{x^{11}}$.
So, $12\sqrt[4]{x}\cdot\sqrt[3]{x^2}=12\sqrt[12]{x^{11}}$.

Answer:

$12\sqrt[12]{x^{11}}$