8th Class Mathematics Cubes and Cube Roots

  • 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.


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