Download the free PDF of RD Sharma Solutions for Class 8 Maths from the links provided below. Students can refer to the RD Sharma textbook and then practise the RD Sharma Class 8 solutions provided by BYJU’S expert team. The solutions are solved using simple techniques for easy understanding, which will help students in achieving high marks in the final examination. In Exercise 4.3 of RD Sharma Solutions for Class 8 Maths Chapter 4 Cubes and Cube Roots, we will discuss problems based on cube roots, the cube root of a natural number using the units digit method and the cube root of a perfect cube by factors.
RD Sharma Solutions for Class 8 Maths Exercise 4.3 Chapter 4 Cubes and Cube Roots
Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 4.3 Chapter 4 Cubes and Cube Roots
1. Find the cube roots of the following numbers by successive subtraction of numbers:
1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, …
(i) 64
(ii) 512
(iii) 1728
Solution:
(i) 64
Let’s perform subtraction
64 – 1 = 63
63 – 7 = 56
56 – 19 =37
37 – 37 = 0
Subtraction is performed 4 times.
∴ The cube root of 64 is 4.
(ii) 512
Let’s perform 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
Subtraction is performed 8 times.
∴ The cube root of 512 is 8.
(iii) 1728
Let’s perform 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
Subtraction is performed 12 times.
∴ The cube root of 1728 is 12.
2. Using the method of successive subtraction, examine whether or not the following numbers are perfect cubes:
(i) 130
(ii) 345
(iii) 792
(iv) 1331
Solution:
(i) 130
Let’s perform subtraction
130 – 1 = 129
129 – 7 = 122
122 – 19 = 103
103 – 37 = 66
66 – 61 = 5
The next number to be subtracted is 91, which is greater than 5
∴130 is not a perfect cube.
(ii) 345
Let’s perform 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 to be subtracted is 169, which is greater than 2
∴ 345 is not a perfect cube.
(iii) 792
Let’s perform 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 to be subtracted is 271, which is greater than 63
∴ 792 is not a perfect cube.
(iv) 1331
Let’s perform 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
Subtraction is performed 11 times
The cube root of 1331 is 11
∴ 1331 is a perfect cube.
3. Find the smallest number that must be subtracted from those of the numbers in question 2, which are not perfect cubes, to make them perfect cubes. What are the corresponding cube roots?
Solution:
In the previous question, there are three numbers which are not perfect cubes.
(i) 130
Let’s perform subtraction
130 – 1 = 129
129 – 7 = 122
122 – 19 = 103
103 – 37 = 66
66 – 61 = 5
The next number to be subtracted is 91, which is greater than 5
Since 130 is not a perfect cube. So, to make it a perfect cube, we subtract 5 from the given number.
130 – 5 = 125 (which is a perfect cube of 5)
(ii) 345
Let’s perform 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 to be subtracted is 169, which is greater than 2
Since 345 is not a perfect cube. So, to make it a perfect cube, we subtract 2 from the given number.
345 – 2 = 343 (which is a perfect cube of 7)
(iii) 792
Let’s perform 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 to be subtracted is 271, which is greater than 63
Since 792 is not a perfect cube. So, to make it a perfect cube, we subtract 63 from the given number.
792 – 63 = 729 (which is a perfect cube of 9)
4. Find the cube root of each of the following natural numbers:
(i) 343 (ii) 2744
(iii) 4913 (iv) 1728
(v) 35937 (vi) 17576
(vii) 134217728 (viii) 48228544
(ix) 74088000 (x) 157464
(xi) 1157625 (xii) 33698267
Solution:
(i) 343
By using the prime factorisation method
∛343 = ∛ (7×7×7) = 7
(ii) 2744
By using the prime factorisation method
∛2744 = ∛ (2×2×2×7×7×7) = ∛ (23×73) = 2×7 = 14
(iii) 4913
By using the prime factorisation method,
∛4913 = ∛ (17×17×17) = 17
(iv) 1728
By using the prime factorisation method,
∛1728 = ∛(2×2×2×2×2×2×3×3×3) = ∛ (23×23×33) = 2×2×3 = 12
(v) 35937
By using the prime factorisation method,
∛35937 = ∛ (3×3×3×11×11×11) = ∛ (33×113) = 3×11 = 33
(vi) 17576
By using the prime factorisation method,
∛17576 = ∛ (2×2×2×13×13×13) = ∛ (23×133) = 2×13 = 26
(vii) 134217728
By using the prime factorisation method
∛134217728 = ∛ (227) = 29 = 512
(viii) 48228544
By using the prime factorisation method
∛48228544 = ∛ (2×2×2×2×2×2×7×7×7×13×13×13) = ∛ (23×23×73×133) = 2×2×7×13 = 364
(ix) 74088000
By using the prime factorisation method
∛74088000 = ∛ (2×2×2×2×2×2×3×3×3×5×5×5×7×7×7) = ∛ (23×23×33×53×73) = 2×2×3×5×7 = 420
(x) 157464
By using the prime factorisation method
∛157464 = ∛ (2×2×2×3×3×3×3×3×3×3×3×3) = ∛ (23×33×33×33) = 2×3×3×3 = 54
(xi) 1157625
By using the prime factorisation method
∛1157625 = ∛ (3×3×3×5×5×5×7×7×7) = ∛ (33×53×73) = 3×5×7 = 105
(xii) 33698267
By using the prime factorisation method
∛33698267 = ∛ (17×17×17×19×19×19) = ∛ (173×193) = 17×19 = 323
5. Find the smallest number, which, when multiplied by 3600, will make the product a perfect cube. Further, find the cube root of the product.
Solution:
Firstly let’s find the prime factors for 3600
3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
= 23 × 32 × 52 × 2
Since only one triple is formed and three factors remain ungrouped in triples.
The given number 3600 is not a perfect cube.
To make it a perfect cube, we have to multiply it by (2 × 2 × 3 × 5) = 60
3600 × 60 = 216000
The cube root of 216000 is
∛216000 = ∛ (60×60×60) = ∛ (603) = 60
∴ The smallest number, which, when multiplied by 3600, will make the product a perfect cube, is 60, and the cube root of the product is 60.
6. Multiply 210125 by the smallest number so that the product is a perfect cube. Also, find out the cube root of the product.
Solution:
The prime factors of 210125 are
210125 = 5 × 5 × 5 × 41 × 41
Since one triple remained incomplete, 210125 is not a perfect cube.
To make it a perfect cube, we need to multiply the factors by 41, we will get 2 triples as 23 and 413.
And the product becomes:
210125 × 41 = 8615125
8615125 = 5 × 5 × 5 × 41 × 41 × 41
Cube root of product = ∛8615125 = ∛ (5×41) = 205
7. What is the smallest number by which 8192 must be divided so that the quotient is a perfect cube? Also, find the cube root of the quotient so obtained.
Solution:
The prime factors of 8192 are
8192 = 2×2×2×2×2×2×2×2×2×2×2 = 23×23×23×2
Since one triple remain incomplete, hence 8192 is not a perfect cube.
So, we divide 8192 by 2 to make its quotient a perfect cube.
8192/2 = 4096
4096 = 2×2×2×2×2×2×2×2×2×2×2×2 = 23×23×23×23
Cube root of 4096 = ∛4096 = ∛ (23×23×23×23) = 2×2×2×2 = 16
8. Three numbers are in the ratio 1:2:3. The sum of their cubes is 98784. Find the numbers.
Solution:
Let us consider the ratio 1:2:3 as x, 2x and 3x
According to the question,
X3 + (2x) 3 + (3x) 3 = 98784
x3 + 8x3 + 27x3 = 98784
36x3 = 98784
x3 = 98784/36
= 2744
x = ∛2744 = ∛ (2×2×2×7×7×7) = 2×7 = 14
So, the numbers are,
x = 14
2x = 2 × 14 = 28
3x = 3 × 14 = 42
9. The volume of a cube is 9261000 m3. Find the side of the cube.
Given volume of cube = 9261000 m3
Let us consider the side of the cube to be ‘a’ metre
So, a3 = 9261000
a = ∛9261000 = ∛ (2×2×2×3×3×3×5×5×5×7×7×7) = ∛ (23×33×53×73) = 2×3×5×7 = 210
∴ the side of cube = 210 metre.
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