Question

9 what is the reduced radical form of (sqrt3{dfrac{243m^4}{3m}})? assume all variables are positive.

Explanation:

Step 1: Simplify the fraction inside the radical

First, simplify \(\frac{243m^{4}}{3m}\). Divide the coefficients: \(243\div3 = 81\). For the variables, use the quotient rule of exponents \(a^{m}\div a^{n}=a^{m - n}\), so \(m^{4}\div m=m^{4 - 1}=m^{3}\). So the fraction simplifies to \(81m^{3}\). Now we have \(\sqrt[3]{81m^{3}}\).

Step 2: Factor the radicand into perfect cubes and other factors

Factor \(81\) as \(27\times3\) (since \(27 = 3^{3}\) is a perfect cube) and \(m^{3}\) is already a perfect cube. So we can rewrite the radicand as \(27\times3\times m^{3}\), which is \(27m^{3}\times3\).

Step 3: Apply the cube - root property

Using the property \(\sqrt[3]{ab}=\sqrt[3]{a}\times\sqrt[3]{b}\) (for \(a,b\geq0\)), we have \(\sqrt[3]{27m^{3}\times3}=\sqrt[3]{27m^{3}}\times\sqrt[3]{3}\). Since \(\sqrt[3]{27m^{3}}=\sqrt[3]{3^{3}m^{3}} = 3m\) (because \(\sqrt[3]{x^{3}}=x\) for \(x\geq0\)), then the expression becomes \(3m\times\sqrt[3]{3}\).

Answer:

\(3m\sqrt[3]{3}\)