Chapter 07 CUBES AND CUBE ROOTS
Exercise 7.1
Q.1 Which of the following numbers are not perfect cubes?
(i) 216 (ii) 128 (iii) 1000 (iv) 100
(v) 46656
Sol. (i) 216
2 | 216 |
2 | 108 |
2 | 54 |
3 | 27 |
3 | 9 |
3 | 3 |
1 |
Prime factors of 216 = 2 x 2 x 2 x 3 x 3 x 3
Here, each prime factor is appearing in multiple of 3.
Hence, 216 is a perfect cube.
(ii) 128
2 | 128 |
2 | 64 |
2 | 32 |
2 | 16 |
2 | 8 |
2 | 4 |
2 | 2 |
1 |
Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
Here, each prime factor is not appearing in multiple of 3.
Hence, 128 is not a perfect cube.
(iii) 1000
2 | 1000 |
2 | 500 |
2 | 250 |
5 | 125 |
5 | 25 |
5 | 5 |
1 |
Prime factors of 1000 = 2 x 2 x 2 x 5 x 5 x 5
Here, each prime factor is appearing in multiple of 3.
Hence, 1000 is a perfect cube.
(iv) 100
2 | 100 |
2 | 50 |
5 | 25 |
5 | 5 |
1 |
Prime factors of 1000 = 2 x 2 x 5 x 5
Here, each prime factor is not appearing in multiple of 3.
Hence, 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 |
Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
Here, each prime factor is appearing in multiple of 3.
Hence, 46656 is a perfect cube.
Q.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) 6750
(v) 100
Sol. (i) 243
3 | 243 |
3 | 81 |
3 | 27 |
3 | 9 |
3 | 3 |
1 |
Prime factors of 243 = 3 x 3 x 3 x 3 x 3
Here, prime factor â€˜3â€™ is not appearing in triplet. Hence, to make 243 a cube, one more â€˜3â€™ needs to be multiplied.
Therefore, 243 x 3 = 3 x 3 x 3 x 3 x 3 x 3 = 729 is a perfect cube.
Hence, the smallest number by which 243 must be multiplied to obtain a perfect cube is 3.
(ii) 256
2 | 256 |
2 | 128 |
2 | 64 |
2 | 32 |
2 | 16 |
2 | 8 |
2 | 4 |
2 | 2 |
1 |
Prime factors of 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
Here, prime factor â€˜2â€™ is not appearing in triplet. Hence, to make 243 a cube, one more â€˜2â€™ needs to be multiplied.
Therefore, 256 x 2 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 512 is a perfect cube.
Hence, the smallest number by which 256 must be multiplied to obtain a perfect cube is 2.
(iii) 72
2 | 72 |
2 | 36 |
2 | 18 |
3 | 9 |
3 | 3 |
1 |
Prime factors of 72 = 2 x 2 x 2 x 3 x 3
Here, prime factor â€˜3â€™ is not appearing in triplet. Hence, to make 72 a cube, one more â€˜3â€™ needs to be multiplied.
Therefore, 72 x 3 = 2 x 2 x 2 x 3 x 3 x 3 = 216 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 |
Prime factors of 675 = 3 x 3 x 3 x 5 x 5
Here, prime factor â€˜5â€™ is not appearing in triplet. Hence, to make 675 a cube, one more â€˜5â€™ needs to be multiplied.
Therefore, 675 x 5 = 3 x 3 x 3 x 5 x 5 x 5 = 3375 is a perfect cube.
Hence, the smallest number by which 675 must be multiplied to obtain a perfect cube is 5.
(v) 100
2 | 100 |
2 | 50 |
5 | 25 |
5 | 5 |
1 |
Prime factors of 100 = 2 x 2 x 5 x 5
Here, prime factor â€˜2â€™ and â€˜5â€™ are not appearing in triplet. Hence, to make 100 a cube, one more â€˜2â€™ and â€˜5â€™ needs to be multiplied.
Therefore, 100 x 2 x 5 = 2 x 2 x 2 x 5 x 5 x 5 = 1000 is a perfect cube.
Hence, the smallest number by which 100 must be multiplied to obtain a perfect cube is 2 x 5 = 10.
Q.3 Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.
(i) 81 (ii) 128 (iii) 135 (iv) 192
(v) 704
Sol. (i) 81
3 | 81 |
3 | 27 |
3 | 9 |
3 | 3 |
1 |
Prime factors of 81 = 3 x 3 x 3 x 3
Here, prime factor â€˜3â€™ is not appearing in triplet. Hence, to make 243 a cube, one â€˜3â€™ needs to be divided.
Therefore, 81 Ã· 3 = 3 x 3 x 3 = 27 is a perfect cube.
Hence, the smallest number by which 81 must be divided to obtain a perfect cube is 3.
(ii) 128
2 | 128 |
2 | 64 |
2 | 32 |
2 | 16 |
2 | 8 |
2 | 4 |
2 | 2 |
1 |
Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
Here, prime factor â€˜2â€™ is not appearing in triplet. Hence, to make 128 a cube, one â€˜2â€™ needs to be divided.
Therefore, 128 Ã· 2 = 2 x 2 x 2 x 2 x 2 x 2 = 64 is a perfect cube.
Hence, the smallest number by which 128 must be divided to obtain a perfect cube is 2.
(iii) 135
3 | 135 |
3 | 45 |
3 | 15 |
5 | 5 |
1 |
Prime factors of 135 = 3 x 3 x 3 x 5
Here, prime factor â€˜5â€™ is not appearing in triplet. Hence, to make 135 a cube, one â€˜5â€™ needs to be divided.
Therefore, 135 Ã· 5 = 3 x 3 x 3 = 27 is a perfect cube.
Hence, the smallest number by which 135 must be divided to obtain a perfect cube is 5.
(iv) 192
2 | 192 |
2 | 96 |
2 | 48 |
2 | 24 |
2 | 12 |
2 | 6 |
3 | 3 |
1 |
Prime factors of 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3
Here, prime factor â€˜3â€™ is not appearing in triplet. Hence, to make 192 a cube, one â€˜3â€™ needs to be divided.
Therefore, 192 Ã· 3 = 2 x 2 x 2 x 2 x 2 x 2 = 64 is a perfect cube.
Hence, the smallest number by which 192 must be divided to obtain a perfect cube is 3.
(v) 704
2 | 704 |
2 | 352 |
2 | 176 |
2 | 88 |
2 | 44 |
2 | 22 |
11 | 11 |
1 |
Prime factors of 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11
Here, prime factor â€˜11â€™ is not appearing in triplet. Hence, to make 704 a cube, one â€˜11â€™ needs to be divided.
Therefore, 704 Ã· 11 = 2 x 2 x 2 x 2 x 2 x 2 = 64 is a perfect cube.
Hence, the smallest number by which 704 must be divided to obtain a perfect cube is 11.
Q.4 Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?
Sol. Given, sides of plasticine are 5 cm, 2 cm, 5 cm.
We know that, volume = 5 cm x 2 cm x 5 cm = (5 x 5 x 2) cm3.
Here, two 5s and one 2 are which are not in a triplet.
Hence, multiplying this by 2 x 2 x 5 = 20, then it will be a perfect cube.
Thus, (5 x 5 x 2 x 2 x 2 x 5) = (5 x 5 x 5 x 2 x 2 x 2) = 1000 is a perfect cube.
Hence, 20 cuboids will be required to form a cube.
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