Question

Find the least 4 digit number greater than 1000 which is a perfect cube​

Answer 1

Hi,

Explanation:

`\sqrt[3]{1000} =10\\\sqrt[3]{9999} =21,5436....\\\\The\ set\ of\ 4\ digit\ number\ perfect\ cube\ greater\ than\ 1000\ is\\\\\{n \in \mathbb{N}\ |\ 1000\leq n^3 \leq 9999\}=\{10,11,12,13,14,15,16,17,18,19,20,21\}`