Question : If $\sqrt[3]{N}$ lies between 6 and 7, where $N$ is an integer then how many values $N$ can take?
Option 1: 126
Option 2: 127
Option 3: 128
Option 4: 125
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Correct Answer: 126
Solution :
Given: If $\sqrt[3]{N}$ lies between 6 and 7, where $N$ is an integer.
$6^3=216$
$7^2=343$
The cube root of 216 is 6 and the cube root of 343 is 7.
Every number between 343 and 216 has a cube root between 6 and 7.
The values $N$ can take = 343 – 216 – 1 = 126.
Hence, the correct answer is 126.
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