Question 1
Write the units digit of the cube for each of the following numbers:(i) 31, (ii) 109, (iii) 388, (iv) 833, (v) 4276, (vi) 5922, (vii) 77774, (viii) 44447, (ix) 125125125
Solution:
(i) 1 is the unit digit of the cube of 31.
(ii) 9 is the unit digit of the cube of 109.
(iii) 2 is the unit digit of the cube of 388.
(iv) 7 is the unit digit of the cube of 833.
(v) 6 is the unit digit of the cube of 4276.
(vi) 8 is the unit digit of the cube of 5922.
(vii) 4 is the unit digit of the cube of 77774.
(viii) 3 is the unit digit of the cube of 44447.
(ix) 5 is the unit digit of the cube of 125125125.
Question 2
Find the cubes of the following numbers:Solution:
(i) 35 = 7 Ã— 5
âˆ´ (35)^{3} = (7)^{3} Ã— (5)^{3 }
Â Â Â Â Â Â Â Â Â Â Â = 7 Ã— 7 Ã— 7 Ã— 5 Ã— 5 Ã— 5
Â Â Â Â Â Â Â Â Â Â Â = 343 Ã— 125
Â Â Â Â Â Â Â Â Â Â Â = 42875.
âˆ´ The cube of 35 = 42875.
(ii) 56 = 7 Ã— 8
âˆ´ (56)^{3} = (7)^{3} Ã— (8)^{3 }
Â Â Â Â Â Â Â Â Â Â Â = 7 Ã— 7 Ã— 7 Ã— 8 Ã— 8 Ã— 8
Â Â Â Â Â Â Â Â Â Â Â = 343 Ã— 512
Â Â Â Â Â Â Â Â Â Â Â = 175616.
âˆ´ The cube of 56= 175616.
(iii) 72 = 8 Ã— 9
âˆ´ (72)^{3} = (8)^{3} Ã— (9)^{3 }
Â Â Â Â Â Â Â Â Â Â Â = 8 Ã— 8 Ã— 8 Ã— 9 Ã— 9 Ã— 9
Â Â Â Â Â Â Â Â Â Â Â = 512 Ã— 729
Â Â Â Â Â Â Â Â Â Â Â = 373248.
âˆ´ The cube of 72 = 373248.
(iv) 402 = 2 Ã— 3 Ã— 67Â
âˆ´ (402)^{3} = (2)^{3} Ã— (3)^{3 }Ã— (67)^{3 }
Â Â Â Â Â Â Â Â Â Â Â = 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 Ã— 67 Ã— 67 Ã— 67Â
Â Â Â Â Â Â Â Â Â Â Â = 8 Ã— 27 Ã— 300763
Â Â Â Â Â Â Â Â Â Â Â = 64964808.
âˆ´ The cube of 402 = 64964808.
(v) 650 = 2 Ã— 5 Ã— 5 Ã— 13Â
âˆ´ (650)^{3} = (2)^{3} Ã— (5)^{3 }Ã— (5)^{3 }Ã— (13)^{3 }
Â Â Â Â Â Â Â Â Â Â Â = 8 Ã— 125 Ã— 125 Ã— 2169
Â Â Â Â Â Â Â Â Â Â Â =274625000.
âˆ´ The cube of 650 = 274625000.
(vi) 819 = 3 Ã— 3 Ã— 7 Ã— 13
âˆ´ (819)^{3} = (3)^{3} Ã— (3)^{3 }Ã— (7)^{3} Ã— (13)^{3 }
Â Â Â Â Â Â Â Â Â Â Â = 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 7 Ã— 7 Ã— 7 Ã— 13 Ã— 13 Ã— 13Â
Â Â Â Â Â Â Â Â Â Â Â = 27 Ã— 27 Ã— 343 Ã—. 2197Â
Â Â Â Â Â Â Â Â Â Â Â = 549353259.
âˆ´ The cube of 819 = 549353259.
Question 3
Which of the following numbers are not perfect cubes?Solution:
(i)Â Â
Â Â Â Â 64Â = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2Â =2^{3} Ã— 2^{3 }= (2 Ã— 2)^{3 Â }Â Â Therefore 64 is a perfect cube.
(ii)Â Â
Â Â Â Â 216Â Â Â Â = 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3=2^{3} Ã— 3^{3 }= (2 Ã— 3)^{3 Â Â Â Â }Therefore^{ }216 is a perfect cube.
(iii)Â Â
Â Â Â Â 243Â Â = 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3=3^{3 }Ã— ^{Â }3^{2}=(3)^{3 }Ã— ^{Â }3^{2}
Â Â Â In the above factorization 3^{2} remains after grouping the 3's in triplets.
Â Â Â Therefore 243 is not a perfect cube.
(iv)Â Â
Â Â Â Â 1728 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 = (2^{3} x 2^{3} x 3^{3}) =(2 Ã— 2 Ã— 3)^{3}
_{Â Â Â Â }1728 is a perfect cube.
Question 4
For each of the nonperfect cubes, find the smallest number by which it must be multiplied so that theÂ product is a perfect cube.Solution:
(i) 243 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2Â
= (3^{3} x 3^{2}) = (3)^{3} x 3^{2} In this factorization there is no triplet for 3. 

(ii) 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = (2^{3} x 2^{3} x 2^{2}) = (2 x 2)^{3} x 2^{2} In this factorization there is no triplet for 2 So 256 is not a perfect cube. âˆ´256 has to be multiplied by 2 to make it a perfect cube. 

(iii) 72 = 2 x 2 x 2 x 3 x 2 x 3 = (2^{3} x 3^{2} x 2) = (2)^{3} x 2 x 3^{2} Â = 12 to make it a perfect cube. 

(iv) 1728 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 = (2^{3 }x 2^{3} x 3^{3}) = (2 x 2 x 3)^{3} Cube root of 1728 is 12.

Question 5
For each of the nonperfect cubes, find the smallest number by which it must be divided so that theÂ quotient is a perfect cube.Â Â Â Â Â Â (i) 81 (ii) 128 (iii) 135 (iv) 192
Solution:
81 = 3 x 3 x 3 x 3 = (3^{3} x 3) = (3)^{3} x 3 In this factorization there is no triplet for 3. So 81 is not a perfect cube. 81 must be divided by 3 to make the quotient a perfect cube. 

128 = 2 x 2 x 2 x 2 x 2 x 2 x 2 = (2^{3} x 2^{3} x 2) = (2 x 2)^{3} x 2 In the above factorization there is no triplet for 2. So 128 is not a perfect cube. âˆ´128 must be divided by 2 to make the quotient a perfect cube. 

Â 135 = 5 x 3 x 3 x 3 Â Â Â Â Â Â = (5 x 3^{3}) Â Â Â Â Â Â = (3)^{3} x 5 In the above factorization there is no triplet for 5. So 135 is not a perfect cube. âˆ´135 must be divided by 5 to make the Quotient a perfect cube. 

192 = 2 x 2 x 2 x 2 x 2 x 2 x 3 Â Â Â Â Â = (2^{3} x 2^{3} x 3) Â Â Â Â Â = (2 x 2)^{3} x 3 In the above factorization there is no triplet for 3. So 192 is not a perfect cube. 
Question 6
By taking three different values of n, verify the truth of the following statements:Â Â Â Â Â Â (i) If n is even, then n^{3} is also even.
Â Â Â Â Â Â (ii) If n is odd, then n^{3} is also odd.
Â Â Â Â Â Â (iii) If n leaves remainder 1 when divided by 3, then n^{3} also leaves 1 as remainder when divided by 3.
Â Â Â Â Â Â (iv) If a natural number n is of the form 3p + 2, then n^{3} is also a number of the same type.
Solution:
(i) LetÂ nÂ =Â 4.
âˆ´ n^{3} = 4 Ã— 4 Ã— 4 = 64.
âˆ´ If n is even, n^{3} is also even is a true statement.
(ii) Let n = 3.
âˆ´ n^{3} = 3 Ã— 3 Ã— 3 = 27.
âˆ´ If n is odd, n^{3} is also odd is a true statement.
(iii) Let n = 10. 10 when divided by 3 leaves a remainder 1.
âˆ´ n^{3} = 10 Ã— 10 Ã— 10 = 1000. 1000 when divided by 3 also leaves a remainder 1.
âˆ´ If n leaves a remainder 1 when divided by 3, then n^{3} also leaves 1 as remainder when divided by 3.
(iv) Let (3 Ã— 5) + 2 = 15 + 2 = 17= n where p = 5.
n^{3} = (17)^{3} = 17 Ã— 17 Ã— 17 = 4913
Let 3 q + 2 = 4913
3 q = 4913 â€“ 2 = 4911
âˆ´ q = = 1637
âˆ´ If a natural number n is of the form 3p + 2, then n^{3} is also a number of the same type.
Question 7
Write true (T) of false (F) for the following statements:Â Â Â Â Â Â (ii) 8640 is not a perfect cube.
Â Â Â Â Â Â (iii) No perfect cube can end with exactly two zeros.
Â Â Â Â Â Â (iv) There is no perfect cube which ends in 4.
Â Â Â Â Â Â (v) For an integer a, a^{3} is always greater than a^{2}.
Â Â Â Â Â Â (vi) If a and b are integers such that a^{2}>b^{2}, then a^{3}>b^{3}.
Â Â Â Â Â Â (vii) If a divides b, then a^{3} divides b^{3}.
Â Â Â Â Â Â (viii) If a^{2} ends in 9, then a^{3} ends in 7.
Â Â Â Â Â Â (ix) If a^{2} ends in 5, then a^{3} ends in 25.
Â Â Â Â Â Â (x) If a^{2} ends in an even number of zeros, then a^{3} ends in an odd number of zeros.
Solution:
(i) False, since 392 is not a perfect cube.
(ii) True.
(iii) True.
(iv) False, since the cube of 4 is 64 which ends in 4.
(v) False, since the square of 1 is 1 is equal to the cube of 1. In this case a^{3} is not greater than a^{2}.
(vi) False, since the square of  2 is 4 is greater than the square of 1 i.e. 1, then the cube of â€“2 is â€“8 is not greater than the cube of 1 ie 1.
âˆ´ If a and b are integers such that a^{2}>b^{2}, then a^{3}>b^{3} is not true in all cases.
(vii) True.
(viii) False, if a = 7, then a^{3} ends in 3. So the statement if a^{2} ends in 9, then a^{3} ends in 7 is not true for all cases.
(ix) False, 15 Ã— 15 = 225 and 15 Ã— 15 Ã— 15 = 3375. So the statement a^{2} ends in 5, then a^{3} ends in 25 is not true in all cases.
(x) False. 100 Ã— 100 = 10000 ends in even number of zeros. 100 Ã— 100 Ã— 100 = 1000000 also ends in even number of zeros. So the statement If a^{2} ends in an even number of zeros, then a^{3} ends in an odd number of zeros is not true for all cases.
Question 8
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,..â€¦ :Solution:
(i) 64 â€“ 1 = 63, 63 â€“ 7 = 56, 56 â€“ 19 = 37, 37 â€“ 37 = 0.
â‡’ The remainder zero is got by 4 successive subtractions.
âˆ´ The cube root of 64 = 4.
(ii) 512 â€“ 1 = 511, 511 â€“ 7 = 504, 504 â€“ 19 = 485, 485 â€“ 37 = 448, 448 â€“ 61 = 387 387 â€“ 91 = 296, 296 â€“ 127 = 169, 169 â€“ 169 =0.
â‡’ The remainder zero is got after 8 successive subtractions.
âˆ´ The cube root of 512 is 8.
(iii) 1728 â€“ 1 = 1727, 1727 â€“ 7 = 1720, 1720 â€“ 19 = 1701, 1701 â€“ 37 = 1664, 1664 â€“ 61 = 1603, 1603 â€“ 91 = 1512, 1512 â€“ 127 = 1385, 1385 â€“169 = 1216, 1216 â€“ 217 = 999, 999 â€“ 271 = 728, 728 â€“ 331 = 397, 397 â€“ 397 = 0.
âˆ´ The remainder zero is got after 12 successive subtractions.
âˆ´ 12 is the cube root of 1728.
Question 9
Â Â Â Â Â Â Â Â (i) 130 (ii) 345 (iii) 792 (iv) 1331
Solution:
Â (i) 130 â€“ 1 = 129, 129 â€“ 7 = 122, 122 â€“ 19 = 103, 103 â€“ 37 = 66, 66 â€“ 61 = 5.
â‡’ The remainder got is 5 not zero by successive subtractions.
âˆ´ 130 is not a perfect cube.
(ii) 345 â€“ 1 = 344, 344 â€“ 7 = 337, 337 â€“ 19 = 318, 318 â€“ 37 = 281, 281 â€“ 61 = 220, 220 â€“ 91 = 219, 219  217 = 2. The next number to be subtracted is 331 which is greater than 2. The remainder got is 2, not zero.
âˆ´ 345 is not a perfect cube.
(iii) 792 â€“ 1 = 791, 791 â€“ 7 = 784, 784 â€“ 19 = 765, 765 â€“ 37 = 728, 728 â€“ 61 = 667, 667 â€“ 91 = 576, 576 â€“ 127 = 449, 449 â€“ 169 = 280, 217 = 63. The next number to be subtracted is 331 which is greater than 63. The remainder got is not zero.
âˆ´ 792 is not a perfect cube.
(iv) 1331 â€“ 1 = 1330, 1330 â€“ 7 = 1323, 1323 â€“ 19 = 1304, 1304 â€“ 37 = 1267, 1267 â€“ 61 = 1206, 1206 â€“ 91 = 1115, 1115 â€“ 127 = 998, 998 â€“ 169 = 819, 819  217 = 602, 602 â€“ 271 = 331, 331 â€“ 331 = 0.
âˆ´ The remainder zero is got after 11 successive subtractions.
âˆ´ The cube root of 1331 = 11.
Question 10
Find the smallest number that must be subtracted from those numbers which are not perfect cubesÂ so as to make them perfect cubes. What are the corresponding cube roots?Solution:
(i) 130 â€“ 1 = 129, 129 â€“ 7 = 122, 122 â€“ 19 = 103, 103 â€“ 37 = 66, 66 â€“ 61 = 5.
â‡’ The remainder got is 5.
â‡’ 5 is the number to be subtracted from 130 to make it a perfect cube.
â‡’ 130 â€“ 5 = 125 is the perfect cube.
âˆ´ The corresponding cube = = 5.
(ii) 345 â€“ 1 = 344, 344 â€“ 7 = 337, 337 â€“ 19 = 318, 318 â€“ 37 = 281, 281 â€“ 61 = 220, 220 â€“ 91 = 129,129  127 = 2. The next number to be subtracted is 169 which is greater than 2. The remainder got is not zero.
âˆ´ 345 is not a perfect cube.
Â Â Also the remainder got is 2.
â‡’ 2 is the number to be subtracted from 345 to make it a perfect cube.
â‡’ 345 â€“ 2 = 343 is the perfect cube.
âˆ´ The corresponding cube root = = 7.
(iii) 792 â€“ 1 = 791, 791 â€“ 7 = 784, 784 â€“ 19 = 765, 765 â€“ 37 = 728, 728 â€“ 61 = 667, 667 â€“ 91 = 576, 576 â€“ 127 = 449, 449 â€“ 169 = 280, 280  217 = 63. The next number to be subtracted is 271 which is greater than 63. 63 is the number to be subtracted from 792 to make it a perfect cube.
âˆ´ 792 â€“ 63 = 729 is the perfect cube.
âˆ´ The corresponding cube root = = 9.
(iv) 1331 â€“ 1 = 1330, 1330 â€“ 7 = 1323, 1323 â€“ 19 = 1304, 1304 â€“ 37 = 1267, 1267 â€“ 61 = 1206, 1206 â€“ 91 = 1115, 1115 â€“ 127 = 988, 988 â€“ 169 = 819, 819  217 = 602, 602 â€“ 271 = 331, 331 â€“ 331 = 0.
âˆ´ The remainder zero is got after 11 successive subtractions.
âˆ´ 1331 is a perfect cube.
Question 11
Find the units digits of the cube roots of the following numbers:Solution:
(i) The unitâ€™s digit of 226981 is 1.
Since (1)^{3} = 1, the unitâ€™s digit of the cube root of 226981 is 1.
(ii) The unitâ€™s digit of 13824 is 4.
Since (4)^{3} = 64, the unitâ€™s digit of the cube root of 13824 is 4.
(iii) The unitâ€™s digit of 571787 is 7.
Since (3)^{3} = 27, the unitâ€™s digit of the cube root of 571787 is 3.
(iv) The unitâ€™s digit of 175616 is 6.
Since (6)^{3} = 216, the unitâ€™s digit of the cube root of 175616 is 6.
Question 12
Find the cube roots of the following numbers by finding their units and tens digits:Solution:
(i) The unitâ€™s digit of 389017 is 7, so 3 is the unitâ€™s digit of the cube root.
Strike out the three digits: units, tens and the hundreds digits. The number left is 389.
7^{3} = 343 < 389 < 512 = 8^{3}
â‡’ 7 is the tenâ€™s digit of the cube root of 389017.
âˆ´ The cube root of 389017 is 73.
(ii) The unitâ€™s digit of 91125 is 5, so 5 is the unitâ€™s digit of the cube root.
Strike out the three digits: units, tens and the hundreds digits. The number left is 91.
4^{3} = 64 < 91 < 125 = 5^{3}
â‡’ 4 is the tenâ€™s digit of the cube root of 91125.
âˆ´ The cube root of 91125 is 45.
(iii) The unitâ€™s digit of 110592 is 2, so 8 is the unitâ€™s digit of the cube root.
Strike out the three digits: units, tens and the hundreds digits. The number left is 110.
4^{3} = 64 < 110 < 125 = 5^{3}
â‡’ 4 is the tenâ€™s digit of the cube root of 110592.
âˆ´ The cube root of 110592 is 48.
(iv) The unitâ€™s digit of 46656 is 6, so 6 is the unitâ€™s digit of the cube root.
Strike out the three digits: units, tens and the hundreds digits. The number left is 44.
3^{3} = 27 < 44 < 64 = 4^{3}
â‡’ 3 is the tenâ€™s digit of the cube root of 46656.
âˆ´ The cube root of 46656 is 36.
Question 13
Find the cube roots of the following numbers using prime factorization:Solution:
(i) 250047
âˆ´ 250047 = (3)^{6} Ã— (7)^{3 }
âˆ´ = 3^{2} Ã— 7 = 63.
(ii) 438976
Â Â Â Â Â 438976 = (2)^{6} Ã— (19)^{3 }
= 2^{2} Ã— 19
Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2 Ã— 2 Ã— 19
Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 76.
(iii) 592704
Â Â Â Â Â Â 592704 = (2)^{6} Ã— (3)^{3} Ã— (7)^{3 }
âˆ´ = (2)^{2} Ã— 3 Ã— 7
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 84.
(iv) 614125
Â Â Â â‡’ 614125 = (5)^{3} Ã— (17)^{3 }
âˆ´ = 5 Ã— 17
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 85.
Question 14
Â Â Â Â Â Â Â Â (i) 226981 (ii) â€“ 13824 (iii) 571787 (iv)  175616
Solution:
(i) 226981
The unitâ€™s digit of 226981 is 1, so 1 is the unitâ€™s digit of the cube root.
Strike out the three digits: units, tens and the hundreds digits. The number left is 226.
6^{3} = 216 < 226 < 343 =7^{3}
â‡’ 6 is the tenâ€™s digit of the cube root of 226981.
âˆ´ 61 is the cube root of â€“226981.
(ii) The unitâ€™s digit of 13824 is 4, so 4 is the unitâ€™s digit of the cube root.
Strike out the three digits: units, tens and the hundreds digits. The number left is 13.
2^{3} = 8 < 13 < 27 =3^{3}
â‡’ 2 is the tenâ€™s digit of the cube root of 13824.
âˆ´ 24 is the cube root of 13824.
(iii) The unitâ€™s digit of 571787 is 7, so 3 is the unitâ€™s digit of the cube root.
Strike out the three digits: units, tens and the hundreds digits. The number left is 571.
8^{3} = 512 < 571 < 729 =9^{3}
â‡’ 8 is the tenâ€™s digit of the cube root of 571787.
âˆ´ 83 is the cube root of 571787.
(iv) The unitâ€™s digit of 175616 is 6, so 6 is the unitâ€™s digit of the cube root.
Strike out the three digits: units, tens and the hundreds digits. The number left is 175.
5^{3} = 125 < 175 < 216 =6^{3}
â‡’ 5 is the tenâ€™s digit of the cube root of 226981.
âˆ´ 56 is the cube root of â€“175616.
Question 15
Â Â Â Â Â Â Â Â (i) 3087 (ii) 33275 (iii) 120393
Solution:
(i) 3087 = (3)^{2} Ã— (7)^{3 } In this factorisation, we find there is no triplet for 3. âˆ´ 3087 does not have a cube root. 

âˆ´ The smallest number by which 3087 is to be multiplied so that the product has a cube root is 3. 

(ii) 33275 = (5)^{2} Ã— (11)^{3 } In this factorisation, we find there is no triplet for 5. âˆ´ 33275 does not have a cube root. 

âˆ´ The smallest number by which 33275 is to be multiplied so that the product has a cube root is 5.  
(iii) 120393 = (3)^{3} Ã— 13 Ã— (17)^{3 } In this factorisation, we find there is no triplet for 13 âˆ´ 120393 does not have a cube root. 

âˆ´ The smallest number by which 120393 is to be multiplied so that the product has a cube root = 13 Ã— 13 = 169. 
Question 16
Find the smallest number by which the numbers given below must be divided so that the quotient has a cube root.(i) 3087 (ii) 33275 (iii) 120393
Solution:
(i) 3087 = (3)^{2} Ã— (7)3 In this factorisation, we find there is no triplet for 3. 

âˆ´ The smallest number by which 3087 is to be divided so that the quotient has a cube root = 3 Ã— 3 = 9.  
(ii) 33275 = (5)^{2} Ã— (11)3 In this factorisation we find there is no triplet for 5 

âˆ´ The smallest number by which 33275 is to divided so that the quotient has a cube root = 5 Ã— 5 = 25. 

(iii) 120393 = (3)^{3} Ã— 13 Ã— (17)3 In this factorisation we find there is no triplet for 13. 

âˆ´ The smallest number by which 120393 is to divided so that the quotient has a cube root = 13 x 13 =169 