question_answer 1)

                Which of the following numbers are not per feet cubes? (i) 216                                    (ii) 128                                   (iii) 1000                               (iv) 100 (v) 46656

Answer:

                (i) 216

2	216
2	108
2	54
3	27
3	9
3	3
	1

 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	128
2	64
2	32
2	16
2	8
2	4
2	2
	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	1000
2	500
2	250
5	125
5	25
5	5
	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	100
2	50
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	46656
2	23328
2	11664
2	5832
2	2916
2	1458
3	729
3	243
3	81
3	27
3	9
3	3
	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.