yazzjazzy0778
yazzjazzy0778 yazzjazzy0778
  • 04-02-2022
  • Mathematics
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Paounn
Paounn Paounn
  • 04-02-2022

Answer:

[tex]5\sqrt[3]{25}[/tex]

Step-by-step explanation:

Let's factor to see if it's a perfect cube: 3 is a factor, so let's divide: [tex]1875:3 = 625[/tex]. 625 is no longer divisible by 3, but it is by 5, in particular it's [tex]5^4 = 625[/tex]. At this point we can start taking out of the cube root as much as we can:

[tex]\sqrt[3]{1875} = \sqrt[3]{3\times 5^4} = \sqrt[3]{3\times 5\times 5^3} = 5 \sqrt[3]{3\times5} = 5\sqrt[3]{25}[/tex]

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