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Exercise 4.3 of Chapter 4 in RD Sharma's Class 8 Mathematics textbook builds upon the foundational knowledge of cubes and cube roots established in previous exercises. This section focuses on more advanced applications and problem-solving techniques involving cube roots. Students will engage with a variety of questions that require them to apply their understanding of cube roots to real-world scenarios, work with larger numbers, and solve multi-step problems. The exercise aims to deepen students' comprehension of the relationships between cubes and cube roots, enhance their computational skills, and develop their ability to approach complex mathematical situations with confidence and precision.
Solution:
(i)64
Performing successive subtraction:
64 – 1 = 63
63 – 7 = 56
56 – 19 =37
37 – 37 = 0
Since subtraction is performed 4 times.
Therefore, the cube root of 64 is 4.
(ii)512
Performing successive subtraction:
512 – 1 = 511
511 – 7 = 504
504 – 19 = 485
485 – 37 = 448
448 – 61 = 387
387 – 91 = 296
296 – 127 = 169
169 – 169 = 0
Since subtraction is performed 8 times.
Therefore, the cube root of 512 is 8.
(iii)1728
Performing successive subtraction:
1728 – 1 = 1727
1727 – 7 = 1720
1720 – 19 = 1701
1701 – 37 = 1664
1664 – 91 = 1512
1512 – 127 = 1385
1385 – 169 = 1216
1216 – 217 = 999
999 – 271 = 728
728 – 331 = 397
397 – 397 = 0
Since subtraction is performed 12 times.
Therefore, the cube root of 1728 is 12.
Solution:
(i)130
Performing successive subtraction:
130 – 1 = 129
129 – 7 = 122
122 – 19 = 103
103 – 37 = 66
66 – 61 = 5
Since the next number to be subtracted is 91, which is greater than 5
Therefore,130 is not a perfect cube.
(ii)345
Performing successive subtraction:
345 – 1 = 344
344 – 7 = 337
337 – 19 = 318
318 – 37 = 281
281 – 61 = 220
220 – 91 = 129
129 – 127 = 2
Since the next number to be subtracted is 169, which is greater than 2
Therefore, 345 is not a perfect cube
(iii) 792
Performing successive subtraction:
792 – 1 = 791
791 – 7 = 784
784 – 19 = 765
765 – 37 = 728
728 – 61 = 667
667 – 91 = 576
576 – 127 = 449
449 – 169 = 280
280 – 217 = 63
Since the next number to be subtracted is 271, which is greater than 63
Therefore, 792 is not a perfect cube
(iv)1331
Performing successive subtraction:
1331 – 1 = 1330
1330 – 7 = 1323
1323 – 19 = 1304
1304 – 37 = 1267
1267 – 61 = 1206
1206 – 91 = 1115
1115 – 127 = 988
988 – 169 = 819
819 – 217 = 602
602 – 271 = 331
331 – 331 = 0
Since subtraction is performed 11 times,
Therefore, Cube root of 1331 is 11
Hence, 1331 is a perfect cube.
Solution:
In the previous question, there are three numbers that are not perfect cubes.
(i)130
Performing successive subtraction:
130 – 1 = 129
129 – 7 = 122
122 – 19 = 103
103 – 37 = 66
66 – 61 = 5
The next number which is to be subtracted is 91, which is greater than 5
Since, 130 is not a perfect cube.
Therefore, to make it a perfect cube we have to subtract 5.
130 – 5 = 125
125 is a perfect cube of 5.
(ii)345
Performing successive subtraction:
345 – 1 = 344
344 – 7 = 337
337 – 19 = 318
318 – 37 = 281
281 – 61 = 220
220 – 91 = 129
129 – 127 = 2
The next number which is to be subtracted is 169, which is greater than 2
Since, 345 is not a perfect cube.
Therefore, to make it a perfect cube we have to subtract 2.
345 – 2 = 343
343 is a perfect cube of 7.
(iii)792
Performing successive subtraction:
792 – 1 = 791
791 – 7 = 784
784 – 19 = 765
765 – 37 = 728
728 – 61 = 667
667 – 91 = 576
576 – 127 = 449
449 – 169 = 280
280 – 217 = 63
The next number which is to be subtracted is 271, which is greater than 63
Since, 792 is not a perfect cube.
Therefore, to make it a perfect cube we have to subtract 63.
792 – 63 = 729
729 is a perfect cube of 9.
Solution:
(i)343
By prime factorizing 343, we get
∛343 = ∛ (7 × 7 × 7) = 7
Therefore, the cube root of 343 is 7
(ii)2744
By prime factorizing 2744, we get
∛2744 = ∛ (2 × 2 × 2 × 7 × 7 × 7)
∛2744 = ∛ (23 × 73) = 2 × 7 = 14
Therefore, the cube root of 2744 is 14
(iii)4913
By prime factorizing 4913, we get
∛4913 = ∛ (17 × 17 × 17) = 17
Therefore, the cube root of 4913 is 17
(iv) 1728
By prime factorizing 1728, we get
∛1728 = ∛(2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3)
∛1728 = ∛ (23 × 23 × 33) = 2 × 2 × 3 = 12
Therefore, the cube root of 1728 is 12
(v)35937
By prime factorizing 35937, we get
∛35937 = ∛ (3 × 3 × 3 × 11 × 11 × 11)
∛35937 = ∛ (33 × 113) = 3 × 11 = 33
Therefore, the cube root of 35937 is 33
(vi) 17576
By prime factorizing 17576, we get
∛17576 = ∛ (2 × 2 × 2 × 13 × 13 × 13)
∛17576 = ∛ (23 × 133) = 2 × 13 = 26
Therefore, the cube root of 17576 is 26
(vii) 134217728
By prime factorizing 134217728, we get
∛134217728 = ∛ (227) = 29 = 512
Therefore, the cube root of 134217728 is 512
(viii)48228544
By prime factorizing 48228544, we get
∛48228544 = ∛ (2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 × 7 × 13 × 13 × 13)
∛48228544 = ∛ (23 × 23 × 73 × 133) = 2 × 2 × 7 × 13 = 364
Therefore, the cube root of 48228544 is 364
(ix)74088000
By prime factorizing 74088000, we get
∛74088000 = ∛ (2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5 × 7 × 7 × 7)
∛74088000 = ∛ (23 × 23 × 33 × 53 × 73) = 2 × 2 × 3 × 5 × 7 = 420
Therefore, the cube root of 74088000 is 420
(x)157464
By prime factorizing 157464, we get
∛157464 = ∛ (2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3)
∛157464 = ∛ (23 × 33 × 33 × 33) = 2 × 3 × 3 × 3 = 54
Therefore, the cube root of 157464 is 54
(xi) 1157625
By prime factorizing 1157625, we get
∛1157625 = ∛ (3 × 3 × 3 × 5 × 5 × 5 × 7 × 7 × 7)
∛1157625 = ∛ (33 × 53 × 73) = 3 × 5 × 7 = 105
Therefore, the cube root of1157625 is 105
(xii)33698267
By prime factorizing 33698267, we get
∛33698267 = ∛ (17 × 17 × 17 × 19 × 19 × 19)
∛33698267 = ∛ (173 × 193) = 17 × 19 = 323
Therefore, the cube root of 33698267 is 323
Solution:
By prime factorizing 3600, we get
3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
By forming groups in triplet of equal factors we get, 3600 = (2 × 2 × 2) × (3 × 3) × (5 × 5) × 2
Since, 2, 3 and 5 cannot form a triplet of equal factors.
Therefore, 3600 must be multiplied with 60 (2 × 2 × 3 × 5) to get a perfect cube.
3600 × 60 = 216000
Cube root of 216000 is
∛216000 = ∛ (60 × 60 × 60)
∛216000 = ∛ (603) = 60
Therefore, the smallest number which when multiplied with 3600 makes a perfect cube is 60.
Solution:
By prime factorizing 210125, we get
210125 = 5 × 5 × 5 × 41 × 41
By forming groups in triplet of equal factors we get, 210125 = (5 × 5 × 5) × (41 × 41)
Since, 41 cannot form a triplet of equal factors.
Therefore, 210125 must be multiplied with 41 to get a perfect cube.
210125 × 41 = 8615125
Now, finding the cube root of 8615125
By using the prime factorization method, we get
8615125 = 5 × 5 × 5 × 41 × 41 × 41
Therefore, Cube root of product = ∛8615125 = ∛ (5 × 41) = 205
Solution:
By prime factorizing 8192, we get
8192 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 23 × 23 × 23 × 2
By forming groups in triplet of equal factors we get, 8192 = (2×2×2)×(2×2×2)×(2×2×2)×(2×2×2)×2
Since, 2 cannot form a triplet of equal factors.
Therefore, 8192 must be divided by 2 to get a perfect cube.
8192/2 = 4096
Now, finding the cube root of 4096
By using the prime factorization method, we get
4096 = 2×2×2×2×2×2×2×2×2×2×2×2 = 23×23×23×23
Therefore, Cube root of 4096 = ∛4096 = ∛ (23×23×23×23) = 2×2×2×2 = 16
Solution:
Given, ratio of number is 1:2:3
Therefore, Let the number be x, 2x and 3x
According to the question, sum of their cube is 98784
x3 + (2x)3+ (3x)3 = 98784
x3 + 8x3 + 27x3 = 98784
36x3 = 98784
x3 = 98784/36
x = 2744
x = ∛2744 = ∛ (2 × 2 × 2 × 7 × 7 × 7)
x = 2×7
x = 14
So, the respected numbers are,
x = 14
2x = 2 × 14 = 28
3x = 3 × 14 = 42
Solution:
Given, the volume of cube = 9261000 m3
Let the side of the cube be ‘x’ meter
Therefore, x3 = 9261000
Taking cube root on both the side,
x = ∛9261000 = ∛ (2×2×2×3×3×3×5×5×5×7×7×7) = ∛ (23×33×53×73) = 2×3×5×7 = 210
Hence, the side of cube is 210 meter
Exercise 4.3 of Chapter 4 provides students with a comprehensive set of problems designed to reinforce and expand their understanding of cubes and cube roots. Through this exercise, students practice finding cube roots of larger numbers, solving word problems involving cubical objects, working with fractional and decimal cube roots, and applying cube root concepts to more complex algebraic expressions. The questions encourage students to think critically about the relationships between a cube's volume, surface area, and edge length, as well as how changes in one dimension affect the others. By completing this exercise, students not only enhance their computational skills but also develop a deeper intuition for three-dimensional mathematical relationships, preparing them for more advanced topics in geometry and algebra.
