Math, asked by sam3005, 8 months ago

perfect cube of 63000

Answers

Answered by sanjeevkumar20056

0

Answer:

Answer:

The principal, real, root of:

63000√3

=39.7905721

All roots:

39.7905721

−19.895286+34.4596463i

−19.895286−34.4596463i

63000 is not a perfect cube

Answered by rajbirsinghtarar

0

Answer:

40

Step-by-step explanation:

40*40*40, 63000 ihope it is brilliant answer

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