Question

simplify: $sqrt3{192c^7}$
options:
$circ$ $8c^3sqrt3{3c}$
$circ$ $4sqrt3{3c^7}$
$circ$ $4c^2sqrt3{3c}$
$circ$ $24c^3sqrt3{3c}$
the movement of the progress bar may be uneven because questions can be worth more or less (including zero) depending on your ans

Explanation:

Step1: Factor the radicand

First, factor \(192c^7\) into perfect - cube factors and non - perfect - cube factors. We know that \(192 = 64\times3=4^{3}\times3\) and \(c^{7}=c^{6}\times c=(c^{2})^{3}\times c\). So, \(\sqrt[3]{192c^{7}}=\sqrt[3]{4^{3}\times3\times(c^{2})^{3}\times c}\).

Step2: Use the property of cube roots

The property of cube roots is \(\sqrt[3]{ab}=\sqrt[3]{a}\times\sqrt[3]{b}\) (where \(a\) and \(b\) are real numbers) and \(\sqrt[3]{x^{3}} = x\).
Applying this property, we have:
\(\sqrt[3]{4^{3}\times3\times(c^{2})^{3}\times c}=\sqrt[3]{4^{3}}\times\sqrt[3]{(c^{2})^{3}}\times\sqrt[3]{3c}\)
Since \(\sqrt[3]{4^{3}} = 4\) and \(\sqrt[3]{(c^{2})^{3}}=c^{2}\), then \(\sqrt[3]{4^{3}}\times\sqrt[3]{(c^{2})^{3}}\times\sqrt[3]{3c}=4c^{2}\sqrt[3]{3c}\)

Answer:

\(4c^{2}\sqrt[3]{3c}\) (corresponding to the option " \(4c^{2}\sqrt[3]{3c}\)")