1188 = 2 x 2 x 3 x 3 x 3 x 11
The primes 2 and 11 do not appear in groups of three. So, 1188 is not a perfect cube. In the factorisation of 1188 the prime 2 appears only two times and the prime 11 appears once. So, if we divide 1188 by 2 x 2 x 11 = 44, then the prime factorisation of the quotient will not contain 2 and 11.
Hence the smallest natural number by which 1188 should be divided to make it a perfect cube is 44.

392 = 2 x 2 x 2 x 7 x 7
The prime factor 7 does not appear in a group of three. Therefore, 392 is not a perfect cube. To make its a cube, we need one more 7. In that case
392 x 7 = 2 x 2 x 2 x 7 x 7 x 7 = 2744
Hence the smallest natural number by which 392 should be multiplied to make a perfect cube is 7.

(a) 1331
(e) 3375
(f) 4913

By prime factorisation, we have

     

6

2.4 m

21

-0.8

27

Given, number is 150.
Now, prime factorisation of 150
150 = 2 x 3 x 5 x 5
The prime factors of 150 do not appear in group of triples.
Hence, 150 is not a perfect cube.

By prime factorisation,
we have
175616 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 7x 7 x 7 = 2 x 2 x 2 x 7 = 56

Thus, the cube root of 175616 is 56.

7³ - 6³ = 1+ 7 x 6 x 3 = 1+ 7 x 18 = 1+ 126 = 127

We have, 53
One's digit of 53 = 3
Now, cube of one's digit of 53 = (3)³ = 3 x 3 x 3 = 27
Hence, the one's digit in the cube of 53 is 7.

We have, 8888
One's digit of 8888 = 8
Now, cube of the one's digit of 8888 = (8)³ = 8 x 8 x 8 =512
Hence,one's digit in the cube of 8888 is 2.

We have, 2744
Resolving 2744 into prime factors, We get
2744= 2 x 2 x 2 x 7 x 7 x 7 =(2 x 7)³=(14)³
Clearly, prime factor of2744 are grouped into triplets
So,2744 is a perfect cube of 14.