Answer: Choice B
![2k^5\sqrt[3]{25}](https://img.qammunity.org/2023/formulas/mathematics/college/8dqpobnv5e8biy2ohp3kxuirvbb66wqw1d.png)
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Work Shown:
![\sqrt[3]{200k^(15)}\\\\\sqrt[3]{8*25k^(5*3)}\\\\\sqrt[3]{8*25(k^5)^3}\\\\\sqrt[3]{8(k^5)^3*25}\\\\\sqrt[3]{8(k^5)^3}*\sqrt[3]{25}\\\\2k^5\sqrt[3]{25}](https://img.qammunity.org/2023/formulas/mathematics/college/a195jpz7akewjkyclwy4oukhbjg43lysic.png)
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Step-by-step explanation:
The goal is to factor the radicand in such a way that we pull out perfect cube factors.
Notice that 200 = 8*25 and 8 is a perfect cube (since 2^3 = 8). It is the largest perfect cube factor of 200.
We can rewrite the k^15 as k^(5*3) which is equivalent to (k^5)^3
Once these perfect cube factors are pulled out, they cancel with the cube root to get what you see above.