Question

simplify the expression: cube root of 96x to the power of 7

Answer 1

To simplify the expression \(\sqrt[3]{96x^7}\), we can break it down into two parts: the numerical part (96) and the variable part (\(x^7\)).

1. **Prime factorization of 96**:
\[
96 = 2^5 \times 3^1
\]

2. **Rewrite the expression**:
Now we can express 96 in its factored form:
\[
\sqrt[3]{96x^7} = \sqrt[3]{2^5 \times 3^1 \times x^7}
\]

3. **Apply the cube root**:
We can separate the cube roots:
\[
\sqrt[3]{2^5} \times \sqrt[3]{3^1} \times \sqrt[3]{x^7}
\]

4. **Simplify each term**:
- For \(2^5\):
\[
\sqrt[3]{2^5} = 2^{5/3} = 2^{1 + 2/3} = 2 \cdot 2^{2/3} = 2 \sqrt[3]{4}
\]

- For \(3^1\):
\[
\sqrt[3]{3^1} = 3^{1/3}
\]

- For \(x^7\):
\[
\sqrt[3]{x^7} = x^{7/3} = x^{2 + 1/3} = x^2 \cdot x^{1/3}
\]

5. **Combine the terms**:
Putting it all together, we have:
\[
\sqrt[3]{96x^7} = 2 \sqrt[3]{4} \cdot 3^{1/3} \cdot x^2 \cdot x^{1/3}
\]

Rearranging this gives:
\[
= 2 \cdot 3^{1/3} \cdot x^2 \cdot \sqrt[3]{4} \cdot x^{1/3}
\]

We can also combine the \(x\) terms:
\[
= 2 \cdot 3^{1/3} \cdot x^2 \cdot x^{1/3} \sqrt[3]{4} = 2 \cdot 3^{1/3} \cdot \sqrt[3]{4} \cdot x^{2 + 1/3} = 2 \cdot 3^{1/3} \cdot \sqrt[3]{4} \cdot x^{7/3}
\]

Thus, the simplified expression is:
\[
2 \cdot 3^{1/3} \sqrt[3]{4} \cdot x^{7/3}
\]