Solution:
A number is a perfect cube only when each factor in the prime factorization is grouped in triples. Using this concept, the smallest number can be identified.
[i] 81
81 = 3 × 3 × 3 × 3
= 33 × 3
Here, the prime factor 3 is not grouped as a triplet. Hence, we divide 81 by 3, so that the obtained number becomes a perfect cube.
Thus, 81 ÷ 3 = 27 = 33 is a perfect cube.
Hence the smallest number by which 81 should be divided to make a perfect cube is 3.
[ii] 128
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
= 23 × 23 × 2
Here, the prime factor 2 is not grouped as a triplet. Hence, we divide 128 by 2, so that the obtained number becomes a perfect cube.
Thus, 128 ÷ 2 = 64 = 43 is a perfect cube.
Hence the smallest number by which 128 should be divided to make a perfect cube is 2.
[iii] 135
135 = 3 × 3 × 3 × 5
= 33 × 5
Here, the prime factor 5 is not a triplet. Hence, we divide 135 by 5, so that the obtained number becomes a perfect cube.
135 ÷ 5 = 27 = 33 is a perfect cube.
Hence the smallest number by which 135 should be divided to make a perfect cube is 5.
[iv] 192
192 = 2 × 2 × 2 × 2 × 2 × 2 × 3
= 23 × 23 × 3
Here, the prime factor 3 is not grouped as a triplet. Hence, we divide 192 by 3, so that the obtained number becomes a perfect cube.
192 ÷ 3 = 64 = 43 is a perfect cube
Hence the smallest number by which 192 should be divided to make a perfect cube is 3.
[v] 704
704 = 2 × 2 × 2 × 2 × 2 × 2 × 11
= 23 × 23 × 11
Here, the prime factor 11 is not grouped as a triplet. Hence, we divide 704 by 11, so that the obtained number becomes a perfect cube.
Thus, 704 ÷ 11 = 64 = 43 is a perfect cube
Hence the smallest number by which 704 should be divided to make a perfect cube is 11.
☛ Check: NCERT Solutions for Class 8 Maths Chapter 7
Video Solution:
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
NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.1 Question 3
Summary:
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 are [i] 3, [ii] 2, [iii] 5, [iv] 3, and [v] 11
☛ Related Questions:
- Which of the following numbers are not perfect cubes?[i] 216 [ii] 128 [iii] 1000 [iv] 100 [v] 46656
- 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
- 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?
- Find the cube root of each of the following numbers by prime factorization method. [i] 64 [ii] 512 [iii] 10648 [iv] 27000 [v] 15625 [vi] 13824 [vii] 110592 [viii] 46656 [ix] 175616 [x] 91125
[i] We have,
1536 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
After grouping the prime factors in triplets, it’s seen that one factor 3 is left without grouping.
1536 = [2 × 2 × 2] × [2 × 2 × 2] × [2 × 2 × 2] × 3
So, in order to make it a perfect cube, it must be divided by 3.
Thus, the smallest number by which 1536 must be divided to obtain a perfect cube is 3.
[ii] We have,
10985 = 5 × 13 × 13 × 13
After grouping the prime factors in triplet, it’s seen that one factor 5 is left without grouping.
10985 = 5 × [13 × 13 × 13]
So, it must be divided by 5 in order to get a perfect cube.
Thus, the required smallest number is 5.
[iii] We have,
28672 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 7
After grouping the prime factors in triplets, it’s seen that one factor 7 is left without grouping.
28672 = [2 × 2 × 2] × [2 × 2 × 2] × [2 × 2 × 2] × [2 × 2 × 2] × 7
So, it must be divided by 7 in order to get a perfect cube.
Thus, the required smallest number is 7.
[iv] 13718 = 2 × 19 × 19 × 19
After grouping the prime factors in triplets, it’s seen that one factor 2 is left without grouping.
13718 = 2 × [19 × 19 × 19]
So, it must be divided by 2 in order to get a perfect cube.
Thus, the required smallest number is 2.