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Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.
(i) 81
(ii) 128
(iii) 135
(iv) 192
(v) 704
Asked by Vijay Last Modified
Answer (i) 81 = 3 × 3 × 3 × 3
Here, one 3 is left which is not in a triplet.
If we divide 81 by 3, then it will become a perfect cube.
Thus, 81 ÷ 3 = 27 = 3 × 3 × 3 is a perfect cube.
Hence, the smallest number by which 81 should be divided to make it a perfect cube is 3.
(ii) 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Here, one 2 is left which is not in a triplet.
If we divide 128 by 2, then it will become a perfect cube.
Thus, 128 ÷ 2 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.
Hence, the smallest number by which 128 should be divided to make it a perfect cube is 2.
(iii) 135 = 3 × 3 × 3 × 5
Here, one 5 is left which is not in a triplet.
If we divide 135 by 5, then it will become a perfect cube.
Thus, 135 ÷ 5 = 27 = 3 × 3 × 3 is a perfect cube.
Hence, the smallest number by which 135 should be divided to make it a perfect cube is 5.
(iv) 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3
Here, one 3 is left which is not in a triplet.
If we divide 192 by 3, then it will become a perfect cube.
Thus, 192 ÷ 3 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.
Hence, the smallest number by which 192 should be divided to make it a perfect cube is 3.
(v) 704 = 2 × 2 × 2 × 2 × 2 × 2 × 11
Here, one 11 is left which is not in a triplet.
If we divide 704 by 11, then it will become a perfect cube.
Thus, 704 ÷ 11 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.
Hence, the smallest number by which 704 should be divided to make it a perfect cube is 11.
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