Answer:
(i) 216
By prime factorisation, \[216=\underline{2\times 2\times 2}\times \underline{3\times 3\times 3}\] | grouping the factors in triplets \[={{2}^{3}}\ \times {{3}^{3}}\] | by laws of exponents \[={{(2\times 3)}^{3}}\] | by laws of exponents \[={{6}^{3}}\] which is a perfect cube. Therefore, 216 is a perfect cube. (ii) 128 2 216 2 108 2 54 3 27 3 9 3 3 1
By prime factorisation, \[128=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times 2\] | grouping the factors in triplets \[={{2}^{3}}\times {{2}^{3}}\times 2\] In the above factorisation, 2 remains after grouping the 2's in triplets. Therefore, 128 is not a perfect cube. (iii) 1000 2 128 2 64 2 32 2 16 2 8 2 4 2 2 1
By prime factorisation, \[1000=\underline{2\times 2\times 2}\times \underline{5\times 5\times 5}\] | grouping the factors in triplets \[={{2}^{3}}\times {{5}^{3}}\] | by laws of exponents \[={{(2\times 5)}^{3}}={{10}^{3}},\] | by laws of exponents which is a perfect cube. Therefore, 1000 is a perfect cube. (iv) 100 2 1000 2 500 2 250 5 125 5 25 5 5 1
By prime factorisation \[100=\underline{2\times 2}\times \underline{5\times 5}\] In the above factorisation, \[2\times 2,\text{ }5\times 5\] remains when try to group the/factors in triplets. Therefore, 100 is not a perfect cube. (v) 46656 2 100 2 50 5 25 5 5 1
By prime factorisation, \[46656=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\]\[\times \underline{3\times 3\times 3}\times \underline{3\times 3\times 3}\] | grouping the factors in triplets \[={{2}^{3}}\times {{2}^{3}}\times {{3}^{3}}\times {{3}^{3}}\] \[={{36}^{3}},\] which is a perfect cube. Therefore, 46656 is a perfect cube. 2 46656 2 23328 2 11664 2 5832 2 2916 2 1458 3 729 3 243 3 81 3 27 3 9 3 3 1
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