question_answer
5)
Find the cube root of each of the following numbers by prime factorisation method: (i) 64 (ii) 512 (iii) 10648 (iv) 27000 (u) 15625 (vi) 13824 (vii) 110592 (viii) 46656 (ix) 175616 (x) 91125.
Answer:
(i) 64 2 | 64 |
2 | 32 |
2 | 16 |
2 | 8 |
2 | 4 |
2 | 2 |
| 1 |
Prime factorisation of 64 is \[\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\] | grouping the factors in triplets \[={{2}^{3}}\times {{2}^{3}}={{(2\times 2)}^{3}}={{4}^{3}}\] | by laws of exponents Therefore, \[\sqrt[3]{64}=4\]. (ii) 512 2 | 512 |
2 | 256 |
2 | 128 |
2 | 64 |
2 | 32 |
2 | 16 |
2 | 8 |
2 | 4 |
2 | 2 |
| 1 |
Prime factorisation of 512 is \[\underline{2\times 2\times 2}\,\times \,\underline{2\times 2\times 2}\,\times \,\underline{2\times 2\times 2}\] | grouping the factors in triple \[={{2}^{3}}\,\times {{2}^{3}}\times {{2}^{3}}={{(2\times 2\times 2)}^{3}}={{8}^{3}}\] | by laws of exponent Therefore, \[\sqrt[3]{512}=8\]. (iii) 10648 2 | 10648 |
2 | 6324 |
2 | 2662 |
11 | 1331 |
11 | 121 |
11 | 11 |
| 1 |
Prime factorisation of 10648 is \[\underline{2\times 2\times 2}\times \underline{11\times 11\times 11}\] | grouping the factors in triplets \[={{2}^{3}}\times {{11}^{3}}\] | by laws of exponents Therefore, \[\sqrt[3]{10648}=2\times 11=22\]. (iv) 27000 2 | 27000 |
2 | 13500 |
2 | 6750 |
3 | 3375 |
3 | 1125 |
3 | 375 |
5 | 125 |
5 | 25 |
5 | 5 |
| 1 |
Prime factorisation of 27000 is \[\underline{2\times 2\times 2}\times \underline{3\times 3\times 3}\times \underline{5\times 5\times 5}\] | grouping the factors in triplets \[={{2}^{3}}\times {{3}^{3}}\times {{5}^{3}}\] | by laws of exponents Therefore, \[\sqrt[3]{27000}=2\times 3\times 5=30\]. (v) 15625 5 | 15625 |
5 | 3125 |
5 | 625 |
5 | 125 |
5 | 25 |
5 | 5 |
| 1 |
Prime factorisation of 15625 is \[\underline{5\times 5\times 5}\,\times \underline{5\times 5\times 5}\] | grouping the factors in triplets \[={{5}^{3}}\times {{5}^{3}}={{(5\times 5)}^{3}}={{25}^{3}}\] | by laws of exponents Therefore, \[\sqrt[3]{15625}\,=5\times 5=25\]. (vi) 13824 2 | 13824 |
2 | 6912 |
2 | 3456 |
2 | 1728 |
2 | 864 |
2 | 432 |
2 | 216 |
2 | 108 |
2 | 54 |
3 | 27 |
3 | 9 |
3 | 3 |
| 1 |
Prime factorisation of 13824 is \[\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\,\times \underline{2\times 2\times 2}\] \[\times \underline{3\times 3\times 3}\] | grouping the factors in triplets \[={{2}^{3}}\times {{2}^{3}}\times {{2}^{3}}\times {{3}^{3}}\] \[={{(2\times 2\times 2\times 3)}^{3}}={{24}^{3}}\] | by laws of exponents Therefore, \[\sqrt[3]{13824}=2\times 2\times 2\times 3=24\]. (vii) 110592 2 | 110592 |
2 | 55296 |
2 | 27648 |
2 | 13824 |
2 | 6912 |
2 | 3456 |
2 | 1728 |
2 | 864 |
2 | 432 |
3 | 27 |
3 | 9 |
3 | 3 |
| 1 |
Prime factorisation of 46656 is \[\,\underline{2\times 2\times 2}\,\times \underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\] \[\,\times \underline{2\times 2\times 2}\times \underline{3\times 3\times 3}\] | grouping the factors in triplets \[={{2}^{3}}\times {{2}^{3}}\times {{2}^{3}}\times {{2}^{3}}\times {{3}^{3}}\] \[={{(2\times 2\times 2\times 2\times 3)}^{3}}={{48}^{3}}\] | by laws of exponents Therefore, \[\sqrt[3]{110592}=2\times 2\times 2\times 2\times 3=48\]. (viii) 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 |
Prime factorisation of 46656 is \[\underline{2\times 2\times 2}\,\times \,\underline{2\times 2\times 2}\times \,\underline{3\times 3\times 3}\]\[\times \,\underline{3\times 3\times 3}\] |grouping the factors in triplets \[={{2}^{3}}\times {{2}^{3}}\times {{3}^{3}}\times {{3}^{3}}\] \[={{(2\times 2\times 3\times 3)}^{3}}={{36}^{3}}\] | by laws of exponents Therefore, \[\sqrt[3]{46656}=2\times 2\times 3\times 3=36\]. (ix) 175616 2 | 175616 |
2 | 87808 |
2 | 43904 |
2 | 21952 |
2 | 10976 |
2 | 5488 |
2 | 2744 |
2 | 1372 |
2 | 686 |
7 | 343 |
7 | 49 |
7 | 7 |
| 1 |
Prime factorisation of 175616 is \[\underline{2\times 2\times 2}\,\times \underline{2\times 2\times 2}\,\times \underline{2\times 2\times 2}\]\[\times \underline{7\times 7\times 7}\] |grouping the factors in triplets \[={{2}^{3}}\times {{2}^{3}}\times {{2}^{3}}\times {{7}^{3}}\] \[={{(2\times 2\times 2\times 7)}^{3}}={{56}^{3}}\] |by laws of exponents Therefore, \[\sqrt[3]{175616}=2\times 2\times 2\times 7=56\]. (x) 91125 3 | 91125 |
3 | 30375 |
3 | 10125 |
3 | 3375 |
3 | 1125 |
3 | 375 |
5 | 125 |
5 | 25 |
5 | 5 |
| 1 |
Prime factorisation of 91125 is \[\underline{3\times 3\times 3}\,\times \underline{3\times 3\times 3}\,\times \underline{5\times 5\times 5}\] |grouping the factors in triplets \[={{3}^{3}}\times {{3}^{3}}\times {{5}^{3}}\] \[={{(3\times 3\times 5)}^{3}}{{45}^{3}}\] |by laws of exponents Therefore, \[\sqrt[3]{91125}\,=3\times 3\times 5=45\].