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, \[\sqrt[3]{91125}\,=3\times 3\times 5=45\].