8th Class Mathematics Cubes and Cube Roots

  • question_answer 2)                 Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube. (i) 243                                    (ii) 256                                   (iii) 72                                    (iv) 675 (v) 100

    Answer:

                    (i) 243

    3 243
    3 81
    3 27
    3 9
    3 3
    1
      By prime factorisation, \[243=\underline{3\times 3\times 3}\times 3\times 3\] | grouping the factors in triplets The prime factor 3 does not appear in a group of three. Therefore, 243 is not a perfect cube. To make it a cube, we need one more 3. In that case \[243\times 3=\underline{3\times 3\times 3}\times \underline{3\times 3\times 3}\] = 729, which is a perfect cube. Hence, the smallest number by which 243 should be multiplied to make a perfect cube is 3. (ii) 256
    2 256
    2 128
    2 64
    2 32
    2 16
    2 8
    2 4
    2 2
    1
      By prime factorisation, \[256=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times 2\times 2\]               | grouping the factors in triplets In the above factorisation 2 remains after grouping 2's in triplets. Therefore, 128 is not a perfect cube. To make it a perfect cube, we need one 2's more. In that case, \[256\times 2=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\] \[={{2}^{3}}\times {{2}^{3}}\times {{2}^{3}}\]                     | by laws of exponents \[={{(2\times 2\times 2)}^{3}}\]                                | by laws of exponents \[={{8}^{3}}=512,\] which is a perfect cube. Hence, the smallest number by which 256 must be multiplied to obtain a perfect cube is 2. The resulting perfect cube is \[512=({{8}^{3}})\]. (iii) 72  
    2 72
    2 36
    2 18
    3 9
    3 3
    1
    By prime factorisation, \[72=\underline{2\times 2\times 2}\times 3\times 3\] | grouping the factors in triplets The prime factors 3 does not appear in a group of three. Therefore, 72 is not a perfect cube. To make it a cube, we need one more 3. In that case, \[72\times 3=\underline{2\times 2\times 2}\times \underline{3\times 3\times 3}\] \[={{2}^{3}}\times {{3}^{3}}\]     | by laws of exponents \[={{(2\ \times 3)}^{3}}\]              | by laws of exponents \[={{6}^{3}},\] which is a perfect cube. Hence, the smallest number by which 72 must be multiplied to obtain a perfect cube is 3. (iv) 675  
    3 675
    3 225
    3 75
    5 25
    5 5
    1
    By prime factorisation, \[675=\underline{3\times 3\times 3}\times 5\times 5\] | grouping the factors in triplets The prime factor 5 does not appear in a group of three. Therefore, 675 is not a perfect cube. To make it a cube, we need one more 5. In that case, \[675\times 5=\underline{3\times 3\times 3}\times \underline{5\times 5\times 5}\] \[={{3}^{3}}\times {{5}^{3}}\]     | by laws of exponents \[={{(3\times 5)}^{3}}\]                 | by laws of exponents \[={{15}^{3}},\] which is a perfect cub Hence, the smallest number by which 675 must be multiplied to obtain a perfect cube is 5. The resulting perfect cube is 3375 \[(={{15}^{3}})\] (v) 100
    2 100
    2 50
    5 25
    5 5
    1
      By prime factorisation, \[100=2\times 2\times 5\times 5\] | grouping the factors in triplets The prime factors 2 and 5 do not appear in a group of three. Therefore, 100 is not a perfect cube. To make it a perfect cube, we need one 2 and one 5 more. In that case, \[100\,\times 2\times 5=\underline{2\times 2\times 2}\times \underline{5\times 5\times 5}\] \[={{2}^{3}}\times {{5}^{3}}\]     | by laws of exponents \[={{(2\times 5)}^{3}}\]                 | by laws of exponents \[={{10}^{3}},\] which is a perfect cube. Hence, the smallest number by which 100 must be multiplied to obtain a perfect cube is \[2\times 5=10\]. The resulting perfect cube is \[1000(={{10}^{3}})\].


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