  Question

# Write true (T)I or false (f) for the followiong statements: (i) 392 is a perfect cube. (ii) 8640 is not a perfect cube. (iii) No cube can end with exactly two zeros. (iv) There is no perfect cube which ends in 4. (v) For an integer a,a3 is always greater than a2  (vi) if a and b are integers such that a2>b2,thena3>b3 (vii) If a adivides b, then a3 divides b3 (viii) If a2 ends in 9 then a3 ends in 7. (ix) If a2 ends in 5, then a3 ends in 25. (x) If a2 ends in an even number of zeros. than a3 ends in an odd number of zeros.

Solution

## (i) False ∵392=2×2×2––––––––––×7×7  ∵ Its all factors are not in triplets of equal factors.  (ii) True ∵8640=2×2×2––––––––––×2×2×2––––––––––×3×3×3––––––––––×5 ∵ 5 is left.  (iii) True: A number ending three zeros can be a perfect cube. (iv) False : ∵(4)3=4×4×4=64 Which ends with 4. (v) False: If n is a proper fraction It is not possible (vi) False: it is not true  if a and b are proper fraction. (vii) rue. (viii) False, as a2 ends in 9 then a3 does not necessarily ends in 7 it ends in 3 also. (ix) False, it is not necessarily that a3 ends in 25  it can end also in 75. (x) False : If a2 ends with even zeros then a3 will ends with odd zeros but of multiple of 3.  Suggest corrections  Similar questions
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