Cubes and Cube Roots

We have 1372 = 2 x 2 x 7 x 7 x 7

Since, the prime factor 2 does not appear in a group of triples.  


∴ 1372  is  not a  perfect cube.

Obviously, to make it a perfect cube we need one more 2 as its factor.

i.e.            [1372] x 2 = [2 x 2 x 7 x 7 x 7] x 2

or                2744 = 2 x 2 x 2 x 7 x 7 x 7

which is a perfect cube.

Thus, the required smallest number = 2.

(i) We have 81 = 3 x 3 x 3 x 3


Grouping the prime factors of 81 into triples, we are left with 3.

∴ 81 is  not a perfect cube
Now,   [81] 3= [3 x 3 x 3 x 3] 3
or    27 = 3 x 3 x 3
i.e. 27 is a perfect cube
Thus, the required smallest number is 3

(ii) we have 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2

Grouping the prime factors of 128 into triples, we are left with 2
∴  128 is  not a perfect cube
Now,  [128] 2 = [2 x 2 x 2 x 2 x 2 x 2 x 2]2
or         64 = 2 x 2 x 2 x 2 x 2 x 2
i.e. 64 is a perfect cube
∴  the smallest required number is 2.

(iii) we have 135 =  3 x 3 x 3 x 5

Grouping the prime factors of 135 into triples, we are left over with 5.
∴  135 is not a perfect cube
Now, [135]5 = [ 3 x 3 x 3 x 5] 5
or      27 = 3 x 3 x 3
i.e. 27 is a perfect cube.
Thus, the required smallest number is 5
(iv) We have 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3

Grouping the prime factors of 192 into triples, 3 is left over.
∴  192 is not a perfect cube.
Now,     [192] 3= [2 x 2 x 2 x 2 x 2 x 2 x 3]3
or            64 = 2 x 2 x 2 x 2 x 2 x 2
i.e. 64 is a perfect cube.
Thus,  the required smallest number is 3.
(v) We have 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11

Grouping the prime factors of 704 into triples, 11 is left over
∴  [704]11 = [2 x 2 x 2 x 2 x 2 x 2 x 11]11
or    64 = 2 x 2 x 2 x 2 x 2 x 2
i.e. 64 is a perfect cube
Thus, the required smallest number is 11.

(i) We have 243 = 3 x 3 x 3 x 3 x 3


The prime factor 3 is not a group of three.
∴ 243 is  not a perfect cube.
Now, [243] x 3 = [3 x 3 x 3 x 3 x 3] x 3
or, 729, = 3 x 3 x 3 x 3 x 3 x 3  
Now, 729 becomes a [perfect cube
Thus, the smallest required number to multipkly 243 to make it a perfect cube is 3.

(ii) We have 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

Grouping the prime factors of 256 in triples, we are left over with 2 x 2.
∴ 256 is  not a perfect cube.
Now, [256] x 2 = [2 x 2 x 2 x 2 x 2 x 2 x 2 x 2] x 2
or, 512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
i.e. 512 is a perfect cube.
thus, the required smallest number is 2.

(iii) we  have 72  = 2 x 2 x 2 x 3 x 3

Grouping the prime factors of 72 in triples, we are left  over with 3 x 3
∴ 72 is  not a perfect cube.
Now, [72] x 3 = [2 x 2 x 2 x 3 x 3] x 3
or,     216 = 2 x 2 x 2 x 3 x 3 x 3
i.e. 216 is a perfect  cube
∴ The smallest number required to multiply 72 to make it a perfect cube is 3.

(iv) We have 675 = 3 x 3 x 3 x 5 x 5
Grouping the prime factors of 675 to triples, we are left over with 5 x 5

∴  675 is not a perfect cube.
Now, [675] x 5 = [3 x 3 x 3 x 5 x 5] x 5
Now, 3375  is a  perfect cube
Thus, the smallest required number to multiply 675 such that the new number is a perfect cube is 5.

(v) We have 100 = 2 x 2 x 5 x 5
The prime factor are not in the groups of triples.


∴  100 is not a perfect cube.
Now, [100] x 2 x 5 = [2 x 2 x 5 x 5] x 2 x 5
or,   [100] x 10 = 2 x 2 x 2 x 5 x 5 x 5

1000 = 2 x 2 x 2 x 5 x 5 x 5

Now, 1000 is a perfect cube
Thus, the required smallest number is 10

(i) Separating the given number (1331) into two groups :

1331  1 and 331
∵ 331 end in 1
∴ Unit's digit of the cube root = 1

∵ 1³ = 1 and 
∴ Ten's digit of the cube root = 1

∴ 

(ii)  Separating the given number (4913) in two groups:

4913 4 and 913
Unit's digit:
∵ Unit's digit in 913  is 3
∴ Unit's digit of the cube root = 7

[7³ = 343 : which ends in 3]

Ten's digit:
∵ 1³ = 1, 2³  = 8
and 1 < 4 < 8
i.e. 1³  <  4 < 2³ 
∴ Then ten's digit of the cube root is 1.

∴  

(iii) Separatibng 12167 in two groups:
     1216712 and 167
Unit's digit :
∵ 167 is ending in 7 and cube of a number ending in 3 ends in 7
∴ The unit's digit of the cube root = 3
Ten's digit
∵ 2³  = 8 adn 3³ = 27
Also,  8 < 12 < 27
or,  2³ < 12 < 3²  

∴ The tens digit of the cube root can be 2.

Thus,   

(iv) separating 32768 in two groups:

32768 32 and 786
Unit's digit:
768 will guess the unit's digit in the cube root.
∵ 768 ends in 8.

∴ Unit's digit in the cube root = 2

Ten's digit:
∵ 3³ = 27 and 4³ - 64
Also, 27 < 32 < 64
or, 3³ < 32 < 4³ 

∴ The ten's digit of the cube root = 3

Thus,