Question

topic: squares, cubes, and beyond
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simplify: $sqrt3{192c^7}$
$4sqrt3{3c^7}$
$4c^2sqrt3{3c}$
$8c^3sqrt{3c}$
$24c^3sqrt{3c}$

Explanation:

Step1: Factor the radicand

First, factor 192 and \(c^7\) into parts that are perfect cubes (since we have a cube root, \(\sqrt[3]{\cdot}\)).

For the numerical part: \(192 = 64 \times 3\), and \(64 = 4^3\) (because \(4 \times 4 \times 4 = 64\)).

For the variable part: \(c^7 = c^{6 + 1} = (c^2)^3 \times c\) (using the exponent rule \(a^{m+n}=a^m\times a^n\) and recognizing that \(c^6=(c^2)^3\) since \((c^2)^3 = c^{2\times3}=c^6\)).

So, we can rewrite the original expression \(\sqrt[3]{192c^7}\) as \(\sqrt[3]{64 \times 3 \times (c^2)^3 \times c}\).

Step2: Apply the cube root property

The cube root property states that \(\sqrt[3]{ab}=\sqrt[3]{a}\times\sqrt[3]{b}\) for real numbers \(a\) and \(b\) (when \(a\) and \(b\) are non - negative, which we can assume here for simplicity). Also, \(\sqrt[3]{a^3}=a\) for any real number \(a\).

Using these properties:
\[

$$\begin{align*} \sqrt[3]{64 \times 3 \times (c^2)^3 \times c}&=\sqrt[3]{64}\times\sqrt[3]{(c^2)^3}\times\sqrt[3]{3c}\\ &= 4\times c^2\times\sqrt[3]{3c} \end{align*}$$

\]

Step3: Combine the terms

Multiply the non - radical terms together. So, \(4\times c^2\times\sqrt[3]{3c}=4c^2\sqrt[3]{3c}\).

Answer:

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