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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:

       (i) 35 (ii) 56 (iii) 72 (iv) 402 (v) 650 (vi) 819

Solution:
(i) 35 = 7 × 5

∴ (35)3 = (7)3 × (5)3

            = 7 × 7 × 7 × 5 × 5 × 5

            = 343 × 125

            = 42875.

∴ The cube of 35 = 42875.

(ii) 56 = 7 × 8

∴ (56)3 = (7)3 × (8)3

            = 7 × 7 × 7 × 8 × 8 × 8

            = 343 × 512

            = 175616.

∴ The cube of 56= 175616.

(iii) 72 = 8 × 9

∴ (72)3 = (8)3 × (9)3

            = 8 × 8 × 8 × 9 × 9 × 9

            = 512 × 729

            = 373248.

∴ The cube of 72 = 373248.

(iv) 402 = 2 × 3 × 67 

∴ (402)3 = (2)3 × (3)3 × (67)3

            = 2 × 2 × 2 × 3 × 3 × 3 × 67 × 67 × 67 

            = 8 × 27 × 300763

            = 64964808.

∴ The cube of 402 = 64964808.

(v) 650 = 2 × 5 × 5 × 13 

∴ (650)3 = (2)3 × (5)3 × (5)3 × (13)3

            = 8 × 125 × 125 × 2169

            =274625000.

∴ The cube of 650 = 274625000.


(vi) 819 = 3 × 3 × 7 × 13

∴ (819)3 = (3)3 × (3)3 × (7)3 × (13)3

            = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7 × 7 × 13 × 13 × 13 

            = 27 × 27 × 343 ×. 2197 

            = 549353259.

∴ The cube of 819 = 549353259.

Question 3

Which of the following numbers are not perfect cubes?

       (i) 64 (ii) 216 (iii) 243 (iv) 1728

Solution:
(i)  
     64 = 2 × 2 × 2 × 2 × 2 × 2 =23 × 23 = (2 × 2)3
 
   Therefore 64 is a perfect cube.


(ii)  
     216     = 2 × 2 × 2 × 3 × 3 × 3=23 × 33 = (2 × 3)3
    
Therefore 216 is a perfect cube.


(iii)  
     243   = 3 × 3 × 3 × 3 × 3=33 ×  32=(3)3 ×  32

    In the above factorization 32 remains after grouping the 3's in triplets.
    Therefore 243 is not a perfect cube.

(iv)  
     1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = (23 x 23 x 33) =(2 × 2 × 3)3

    1728 is a perfect cube.


Question 4

For each of the non-perfect cubes, find the smallest number by which it must be multiplied so that the product is a perfect cube.

       (i) 243 (ii) 256 (iii) 72 (iv) 1728

Solution:
(i) 243 = 2 × 2 × 2 × 2 × 2 ×

= (33 x 32) = (3)3 x 32

In this factorization there is no triplet for 3.
So 243 is not a perfect cube.

243 has to be multiplied by 3 to make it a perfect cube.

(ii) 256
= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
= (23 x 23 x 22)
= (2 x 2)3 x 22

In this factorization there is no triplet for 2

So 256 is not a perfect cube.


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 = (23 x 32 x 2) = (2)3 x 2 x 32

In this factorization there is no triplet for 2 and 3

So 72 is not a perfect cube.


72 has to be multiplied by 22 x 31

 = 12 to make it a perfect cube.

(iv) 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = (23 x 23 x 33) = (2 x 2 x 3)3

Cube root of 1728 is 12.
Therefore 1728 is a perfect cube.


Question 5

For each of the non-perfect cubes, find the smallest number by which it must be divided so that the quotient is a perfect cube.
   
    (i) 81 (ii) 128 (iii) 135 (iv) 192

Solution:
81 = 3 x 3 x 3 x 3
= (33 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
= (23 x 23 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.

128 must be divided by 2 to make the
quotient a perfect cube.
 135 = 5 x 3 x 3 x 3
       = (5 x 33)
       = (3)3 x 5
In the above factorization there is no triplet
for 5.
So 135 is not a perfect cube.

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
      = (23 x 23 x 3)
      = (2 x 2)3 x 3
In the above factorization there is no triplet for 3.

So 192 is not a perfect cube.
192 must be divided by 3 to make the Quotient a perfect cube.


Question 6

By taking three different values of n, verify the truth of the following statements:
       (i) If n is even, then n3 is also even.
       (ii) If n is odd, then n3 is also odd.
       (iii) If n leaves remainder 1 when divided by 3, then n3 also leaves 1 as remainder when divided by 3.
       (iv) If a natural number n is of the form 3p + 2, then n3 is also a number of the same type.

Solution:
(i) Let  n  =  4.

∴ n3 = 4 × 4 × 4 = 64.

∴ If n is even, n3 is also even is a true statement.

(ii) Let n = 3.

∴ n3 = 3 × 3 × 3 = 27.

∴ If n is odd, n3 is also odd is a true statement.

(iii) Let n = 10. 10 when divided by 3 leaves a remainder 1.

∴ n3 = 10 × 10 × 10 = 1000. 1000 when divided by 3 also leaves a remainder 1.

∴ If n leaves a remainder 1 when divided by 3, then n3 also leaves 1 as remainder when divided by 3.

(iv) Let (3 × 5) + 2 = 15 + 2 = 17= n where p = 5.

n3 = (17)3 = 17 × 17 × 17 = 4913

Let 3 q + 2 = 4913

3 q = 4913 – 2 = 4911

∴ q = = 1637

∴ If a natural number n is of the form 3p + 2, then n3 is also a number of the same type.

Question 7

Write true (T) of false (F) for the following statements:

       (i) 392 is a perfect cube.
       (ii) 8640 is not a perfect cube.
       (iii) No perfect cube can end with exactly two zeros.
       (iv) There is no perfect cube which ends in 4.
       (v) For an integer a, a3 is always greater than a2.
       (vi) If a and b are integers such that a2>b2, then a3>b3.
       (vii) If a divides b, then a3 divides b3.
       (viii) If a2 ends in 9, then a3 ends in 7.
       (ix) If a2 ends in 5, then a3 ends in 25.
       (x) If a2 ends in an even number of zeros, then a3 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 a3 is not greater than a2.

(vi) False, since the square of - 2 is 4 is greater than the square of 1 i.e. 1, then the cube of –2 is –8 is not greater than the cube of 1 ie 1.

∴ If a and b are integers such that a2>b2, then a3>b3 is not true in all cases.

(vii) True.

(viii) False, if a = 7, then a3 ends in 3. So the statement if a2 ends in 9, then a3 ends in 7 is not true for all cases.

(ix) False, 15 × 15 = 225 and 15 × 15 × 15 = 3375. So the statement a2 ends in 5, then a3 ends in 25 is not true in all cases.

(x) False. 100 × 100 = 10000 ends in even number of zeros. 100 × 100 × 100 = 1000000 also ends in even number of zeros. So the statement If a2 ends in an even number of zeros, then a3 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,..… :

       (i) 64 (ii) 512 (iii) 1728

Solution:
(i) 64 – 1 = 63, 63 – 7 = 56, 56 – 19 = 37, 37 – 37 = 0.

The remainder zero is got by 4 successive subtractions.

∴ The cube root of 64 = 4.

(ii) 512 – 1 = 511, 511 – 7 = 504, 504 – 19 = 485, 485 – 37 = 448, 448 – 61 = 387 387 – 91 = 296, 296 – 127 = 169, 169 – 169 =0.

The remainder zero is got after 8 successive subtractions.

∴ The cube root of 512 is 8.

(iii) 1728 – 1 = 1727, 1727 – 7 = 1720, 1720 – 19 = 1701, 1701 – 37 = 1664, 1664 – 61 = 1603, 1603 – 91 = 1512, 1512 – 127 = 1385, 1385 –169 = 1216, 1216 – 217 = 999, 999 – 271 = 728, 728 – 331 = 397, 397 – 397 = 0.

The remainder zero is got after 12 successive subtractions.

∴ 12 is the cube root of 1728.

Question 9

Using the method of successive subtraction of numbers 1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331,397, ……… examine if the following numbers are perfect cubes:
         (i) 130 (ii) 345 (iii) 792 (iv) 1331

Solution:
 (i) 130 – 1 = 129, 129 – 7 = 122, 122 – 19 = 103, 103 – 37 = 66, 66 – 61 = 5.

The remainder got is 5 not zero by successive subtractions.

∴ 130 is not a perfect cube.

(ii) 345 – 1 = 344, 344 – 7 = 337, 337 – 19 = 318, 318 – 37 = 281, 281 – 61 = 220, 220 – 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.

∴ 345 is not a perfect cube.

(iii) 792 – 1 = 791, 791 – 7 = 784, 784 – 19 = 765, 765 – 37 = 728, 728 – 61 = 667, 667 – 91 = 576, 576 – 127 = 449, 449 – 169 = 280, 217 = 63. The next number to be subtracted is 331 which is greater than 63. The remainder got is not zero.

∴ 792 is not a perfect cube.

(iv) 1331 – 1 = 1330, 1330 – 7 = 1323, 1323 – 19 = 1304, 1304 – 37 = 1267, 1267 – 61 = 1206, 1206 – 91 = 1115, 1115 – 127 = 998, 998 – 169 = 819, 819 - 217 = 602, 602 – 271 = 331, 331 – 331 = 0.

∴ The remainder zero is got after 11 successive subtractions.

∴ The cube root of 1331 = 11.

Question 10

Find the smallest number that must be subtracted from those numbers which are not perfect cubes so as to make them perfect cubes. What are the corresponding cube roots?

          (i) 130 (ii) 345 (iii) 792 (iv) 1331

Solution:

(i) 130 – 1 = 129, 129 – 7 = 122, 122 – 19 = 103, 103 – 37 = 66, 66 – 61 = 5.

The remainder got is 5.

5 is the number to be subtracted from 130 to make it a perfect cube.

130 – 5 = 125 is the perfect cube.

∴ The corresponding cube = = 5.

(ii) 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. The remainder got is not zero.

∴ 345 is not a perfect cube.
   Also the remainder got is 2.

2 is the number to be subtracted from 345 to make it a perfect cube.

345 – 2 = 343 is the perfect cube.

∴ The corresponding cube root = = 7.

(iii) 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. 63 is the number to be subtracted from 792 to make it a perfect cube.

∴ 792 – 63 = 729 is the perfect cube.

∴ The corresponding cube root = = 9.

(iv) 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.

∴ The remainder zero is got after 11 successive subtractions.

∴ 1331 is a perfect cube.

Question 11

Find the units digits of the cube roots of the following numbers:

         (i) 226981 (ii) 13824 (iii) 571787 (iv) 175616

Solution:
(i) The unit’s digit of 226981 is 1.

Since (1)3 = 1, the unit’s digit of the cube root of 226981 is 1.

(ii) The unit’s digit of 13824 is 4.

Since (4)3 = 64, the unit’s digit of the cube root of 13824 is 4.

(iii) The unit’s digit of 571787 is 7.

Since (3)3 = 27, the unit’s digit of the cube root of 571787 is 3.

(iv) The unit’s digit of 175616 is 6.

Since (6)3 = 216, the unit’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:

         (i) 389017 (ii) 91125 (iii) 110592 (iv) 46656

Solution:
(i) The unit’s digit of 389017 is 7, so 3 is the unit’s digit of the cube root.

Strike out the three digits: units, tens and the hundreds digits. The number left is 389.

73 = 343 < 389 < 512 = 83

⇒ 7 is the ten’s digit of the cube root of 389017.

∴ The cube root of 389017 is 73.

(ii) The unit’s digit of 91125 is 5, so 5 is the unit’s digit of the cube root.

Strike out the three digits: units, tens and the hundreds digits. The number left is 91.

43 = 64 < 91 < 125 = 53

⇒ 4 is the ten’s digit of the cube root of 91125.

∴ The cube root of 91125 is 45.

(iii) The unit’s digit of 110592 is 2, so 8 is the unit’s digit of the cube root.

Strike out the three digits: units, tens and the hundreds digits. The number left is 110.

43 = 64 < 110 < 125 = 53

⇒ 4 is the ten’s digit of the cube root of 110592.

∴ The cube root of 110592 is 48.

(iv) The unit’s digit of 46656 is 6, so 6 is the unit’s digit of the cube root.

Strike out the three digits: units, tens and the hundreds digits. The number left is 44.

33 = 27 < 44 < 64 = 43

⇒ 3 is the ten’s digit of the cube root of 46656.

∴ The cube root of 46656 is 36.

Question 13

Find the cube roots of the following numbers using prime factorization:

         (i) 250047 (ii) 438976 (iii) 592704 (iv) 614125

Solution:
(i) 250047

∴ 250047 = (3)6 × (7)3

= 32 × 7 = 63.

(ii) 438976

      438976 = (2)6 × (19)3

= 22 × 19

               = 2 × 2 × 19

               = 76.

(iii) 592704

       592704 = (2)6 × (3)3 × (7)3

= (2)2 × 3 × 7

                  = 84.

(iv) 614125

    ⇒ 614125 = (5)3 × (17)3

= 5 × 17

                  = 85.

Question 14

Find the cube roots of
         (i) -226981 (ii) – 13824 (iii) -571787 (iv) - 175616

Solution:
(i) -226981

The unit’s digit of 226981 is 1, so 1 is the unit’s digit of the cube root.

Strike out the three digits: units, tens and the hundreds digits. The number left is 226.

63 = 216 < 226 < 343 =73

⇒ 6 is the ten’s digit of the cube root of 226981.

∴ -61 is the cube root of –226981.

(ii) The unit’s digit of 13824 is 4, so 4 is the unit’s digit of the cube root.

Strike out the three digits: units, tens and the hundreds digits. The number left is 13.

23 = 8 < 13 < 27 =33

⇒ 2 is the ten’s digit of the cube root of 13824.

∴ -24 is the cube root of -13824.

(iii) The unit’s digit of 571787 is 7, so 3 is the unit’s digit of the cube root.

Strike out the three digits: units, tens and the hundreds digits. The number left is 571.

83 = 512 < 571 < 729 =93

⇒ 8 is the ten’s digit of the cube root of 571787.

∴ -83 is the cube root of -571787.

(iv) The unit’s digit of 175616 is 6, so 6 is the unit’s digit of the cube root.

Strike out the three digits: units, tens and the hundreds digits. The number left is 175.

53 = 125 < 175 < 216 =63

⇒ 5 is the ten’s digit of the cube root of 226981.

∴ -56 is the cube root of –175616.

Question 15

Examine whether or not each of the following numbers has a cube root. If not, find the smallest number by which the number must be multiplied so that the product has a cube root.
         (i) 3087 (ii) 33275 (iii) 120393

Solution:

(i) 3087 = (3)2 × (7)3

In this factorisation, we find there is no triplet for 3.

3087 does not have a cube root.

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

(ii) 33275 = (5)2 × (11)3

In this factorisation, we find there is no triplet for 5.

33275 does not have a cube root.

∴ The smallest number by which 33275 is to be multiplied so that the product has a cube root is 5.

(iii) 120393 = (3)3 × 13 × (17)3

In this factorisation, we find there is no triplet for 13

∴ 120393 does not have a cube root.

∴ The smallest number by which 120393 is to be multiplied so that the product has a cube root = 13 × 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 × (7)3
In this factorisation, we find there is no triplet for 3.
∴ The smallest number by which 3087 is to be divided so that the quotient has a cube root = 3 × 3 = 9.
(ii) 33275 = (5)2 × (11)3

In this factorisation we find there is no triplet for 5

∴ The smallest number by which 33275 is to divided so that the quotient has a cube root = 5 × 5 = 25.

(iii) 120393 = (3)3 × 13 × (17)3

In this factorisation we find there is no triplet for 13.


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




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