Question

simplify the radical expression. (sqrt3{-432x^{9}})

Explanation:

Step1: Factor the number and variable

First, factor \(-432\) and \(x^9\) into parts that are perfect cubes. We know that \(-432 = -8\times54=-8\times27\times2\) and \(x^9=(x^3)^3\). Also, \(-8 = (-2)^3\) and \(27 = 3^3\). So we can rewrite the radicand as:
$$\sqrt[3]{-432x^9}=\sqrt[3]{(-2)^3\times3^3\times2\times(x^3)^3}$$

Step2: Use the property of cube roots

The property of cube roots states that \(\sqrt[3]{ab}=\sqrt[3]{a}\times\sqrt[3]{b}\) (for real numbers \(a\) and \(b\)) and \(\sqrt[3]{a^3}=a\). Applying these properties:
$$\sqrt[3]{(-2)^3\times3^3\times2\times(x^3)^3}=\sqrt[3]{(-2)^3}\times\sqrt[3]{3^3}\times\sqrt[3]{(x^3)^3}\times\sqrt[3]{2}$$
Since \(\sqrt[3]{(-2)^3}=-2\), \(\sqrt[3]{3^3}=3\) and \(\sqrt[3]{(x^3)^3}=x^3\), we substitute these values in:
$$(-2)\times3\times x^3\times\sqrt[3]{2}$$

Step3: Simplify the product

Multiply the constant terms: \(-2\times3=-6\). So the simplified form is:
$$-6x^3\sqrt[3]{2}$$

Answer:

\(-6x^{3}\sqrt[3]{2}\)