Exercise 7.1
Question 1
Which of the following numbers are not perfect cubes?
(i) 216
(ii) 128
(iii) 1000
(iv) 100
(v) 46656
Sol :
(i) The prime factorization of 216 is as follows.

2
216 2
108 2
54 3
27 3
9 3
3 1
216 = 2 × 2 × 2 × 3 × 3 × 3 = 2^{3} × 3^{3}
Here, as each prime factor is appearing as many times as a perfect multiple of 3, therefore, 216 is a perfect cube.
(ii)The prime factorization of 128 is as follows.

2
128 2
64 2
32 2
16 2
8 2
4 2
2 1
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Here, each prime factor is not appearing as many times as a perfect multiple of 3. One 2 is remaining after grouping the triplets of 2. Therefore, 128 is not a perfect cube.
(iii) The prime factorization of 1000 is as follows.

2
1000 2
500 2
250 5
125 5
25 5
5 1
1000 = 2 × 2 × 2 × 5 × 5 × 5
Here, as each prime factor is appearing as many times as a perfect multiple of 3, therefore, 1000 is a perfect cube.
(iv)The prime factorization of 100 is as follows.

2
100 2
50 5
25 5
5 1
100 = 2 × 2 × 5 × 5
Here, each prime factor is not appearing as many times as a perfect multiple of 3. Two 2s and two 5s are remaining after grouping the triplets. Therefore, 100 is not a perfect cube.
(v)The prime factorization of 46656 is as follows.

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
46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
Here, as each prime factor is appearing as many times as a perfect multiple of 3, therefore, 46656 is a perfect cube.
Question 2
Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.
(i) 243
Sol :
(i) 243 = 3 × 3 × 3 × 3 × 3
Here, two 3s are left which are not in a triplet. To make 243 a cube, one more 3 is required.
In that case, 243 × 3 = 3 × 3 × 3 × 3 × 3 × 3 = 729 is a perfect cube.
Hence, the smallest natural number by which 243 should be multiplied to make it a perfect cube is 3.
(ii) 256
Sol :
(ii) 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
Here, two 2s are left which are not in a triplet. To make 256 a cube, one more 2 is required.
Then, we obtain
256 × 2 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512 is a perfect cube.
Hence, the smallest natural number by which 256 should be multiplied to make it a perfect cube is 2.
(iii) 72
Sol :
(iii) 72 = 2 × 2 × 2 × 3 × 3
Here, two 3s are left which are not in a triplet. To make 72 a cube, one more 3 is required.
Then, we obtain
72 × 3 = 2 × 2 × 2 × 3 × 3 × 3 = 216 is a perfect cube.
Hence, the smallest natural number by which 72 should be multiplied to make it a perfect cube is 3.
(iv) 675
Sol :
(iv) 675 = 3 × 3 × 3 × 5 × 5
Here, two 5s are left which are not in a triplet. To make 675 a cube, one more 5 is required.
Then, we obtain
675 × 5 = 3 × 3 × 3 × 5 × 5 × 5 = 3375 is a perfect cube.
Hence, the smallest natural number by which 675 should be multiplied to make it a perfect cube is 5.
(v) 100
Sol :
(v) 100 = 2 × 2 × 5 × 5
Here, two 2s and two 5s are left which are not in a triplet. To make 100 a cube, we require one more 2 and one more 5.
Then, we obtain
100 × 2 × 5 = 2 × 2 × 2 × 5 × 5 × 5 = 1000 is a perfect cube
Hence, the smallest natural number by which 100 should be multiplied to make it a perfect cube is 2 × 5 = 10.
Question 3
Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.
(i) 81
Sol :
(i) 81 = 3 × 3 × 3 × 3
Here, one 3 is left which is not in a triplet.
If we divide 81 by 3, then it will become a perfect cube.
Thus, 81 ÷ 3 = 27 = 3 × 3 × 3 is a perfect cube.
Hence, the smallest number by which 81 should be divided to make it a perfect cube is 3.
(ii) 128
Sol :
(ii) 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Here, one 2 is left which is not in a triplet.
If we divide 128 by 2, then it will become a perfect cube.
Thus, 128 ÷ 2 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.
Hence, the smallest number by which 128 should be divided to make it a perfect cube is 2.
(iii) 135
Sol :
(iii) 135 = 3 × 3 × 3 × 5
Here, one 5 is left which is not in a triplet.
If we divide 135 by 5, then it will become a perfect cube.
Thus, 135 ÷ 5 = 27 = 3 × 3 × 3 is a perfect cube.
Hence, the smallest number by which 135 should be divided to make it a perfect cube is 5.
(iv) 192
Sol :
(iv) 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3
Here, one 3 is left which is not in a triplet.
If we divide 192 by 3, then it will become a perfect cube.
Thus, 192 ÷ 3 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.
Hence, the smallest number by which 192 should be divided to make it a perfect cube is 3.
(v) 704
Sol :
(v) 704 = 2 × 2 × 2 × 2 × 2 × 2 × 11
Here, one 11 is left which is not in a triplet.
If we divide 704 by 11, then it will become a perfect cube.
Thus, 704 ÷ 11 = 64 = 2 × 2 × 2 × 2 × 2 × 2 is a perfect cube.
Hence, the smallest number by which 704 should be divided to make it a perfect cube is 11.
Question 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 :
Here, some cuboids of size 5 × 2 × 5 are given.
When these cuboids are arranged to form a cube, the side of this cube so formed will be a common multiple of the sides (i.e., 5, 2, and 5) of the given cuboid.
LCM of 5, 2, and 5 = 10
Let us try to make a cube of 10 cm side.
For this arrangement, we have to put 2 cuboids along with its length, 5 along with its width, and 2 along with its height.
Total cuboids required according to this arrangement = 2 × 5 × 2 = 20
With the help of 20 cuboids of such measures, a cube is formed as follows.
Alternatively
Volume of the cube of sides 5 cm, 2 cm, 5 cm
= 5 cm × 2 cm × 5 cm = (5 × 5 × 2) cm^{3}
Here, two 5s and one 2 are left which are not in a triplet.
If we multiply this expression by 2 × 2 × 5 = 20, then it will become a perfect cube.
Thus, (5 × 5 × 2 × 2 × 2 × 5) = (5 × 5 × 5 × 2 × 2 × 2) = 1000 is a perfect cube. Hence, 20 cuboids of 5 cm, 2 cm, 5 cm are required to form a cube.
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