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To find the smallest number by which each of the given numbers must be divided to obtain a perfect cube, we need to factorize each number into its prime factors and identify the factors that are not already perfect cubes. Then, we can find the smallest number that needs to be multiplied to these factors to make them perfect cubes.

i. 81 = 3^4 81 is already a perfect cube, so we do not need to divide it by any number.

ii. 128 = 2^7 The factor 2 appears with an odd exponent (7), so we need to divide 128 by 2^1 to get 64, which is a perfect cube: 128 Ã· 2 = 64

iii. 135 = 3^3 x 5^1 Both 3 and 5 appear with odd exponents, so we need to divide 135 by 3^2 x 5^1 to get 3 x 3 x 3 x 5 = 135, which is a perfect cube: 135 Ã· (3^2 x 5) = 3 x 3 x 3 x 5 = 135

iv. 192 = 2^6 x 3^1 The factor 2 appears with an even exponent (6), and the factor 3 appears with an odd exponent (1). We need to divide 192 by 2^2 x 3^2 to get 2^4 x 3 = 48, which is a perfect cube: 192 Ã· (2^2 x 3^2) = 2^4 x 3 = 48

v. 704 = 2^6 x 11^1 The factor 2 appears with an even exponent (6), and the factor 11 appears with an odd exponent (1). We need to divide 704 by 2^2 x 11^1 to get 2^4 x 11 = 176, which is a perfect cube: 704 Ã· (2^2 x 11) = 2^4 x 11 = 176

Therefore, the smallest number by which each of the given numbers must be divided to obtain a perfect cube are: i. 81 (already a perfect cube) ii. 2 (128 Ã· 2 = 2^1 x 64 = 2^3 x 8 = 2^3 x 2^3) iii. 9 (135 Ã· (3^2 x 5^1) = 3 x 3 x 3 x 5 = 3^3 x 5^1) iv. 12 (192 Ã· (2^2 x 3^2) = 2^4 x 3 = 2^2 x 2^2 x 3^3) v. 22 (704 Ã· (2^2 x 11) = 2^4 x 11 = 2^2 x 2^2 x 11^3)