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If x2 < 625, and |x + 8| < 31, then how many values of x are perfect cubes of odd natural numbers?
    Correct answer is '1'. Can you explain this answer?
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    If x2 < 625, and |x + 8| < 31, then how many values of x are per...
    Solution: x2 < 625 that is, -25 < x < 25
    |x + 8| < 31 that is, -39 < x < 23
    So -25 < x < 23 All cubes that come within this range are - 8 , - 1 , 0 , 1 ,8 .
    Out of these only 1 is a perfect cube of an odd natural number.
    Answer: 1
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    If x2 < 625, and |x + 8| < 31, then how many values of x are per...
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    If x2 < 625, and |x + 8| < 31, then how many values of x are perfect cubes of odd natural numbers?Correct answer is '1'. Can you explain this answer?
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