125000, 50

Given, number is 23.
Cube of 23 = 23 x 23 x 23 = 12167

15625

No

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. 
\(\therefore\) [704] \(\div\)11 
= [2 x 2 x 2 x 2 x 2 x 2 x 11]\(\div\) 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

We have 27000000 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 5 x 5 x 5 x 5 x 5 x 5
Since, all the prime factors of 27000000 appear in groups of triples.

\(\therefore\) 27000000 is a perfect cube.

We have
64000 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5 x 5

Since, we get groups of triples.
\(\therefore\)64000 is a perfect cube.

We have, 3375
Resolving 3375 into prime factors, we get
3375 = 3 x 3 x 3 x 5 x 5 x 5
Clearly, the prime factors of 3375 are grouped into triples of equal factors and no factor is left over.
So, 3375 is a perfect cube.

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

From the given pattern, we observe that it follows the relation
n³ = [n (n -1) + 1]+ [n (n -1) + 3] + [n (n -1) + 5] + ...+n terms
6³ = [6 (6 -1) + 1]+ [6 (6 -1) + 3) + [6 (6 -1) + 5]+ [6 (6 -1) + 7) + [6 (6 -1) + 9] + [6 (6 -1) + 11]
=[6 x 5 + 1]+ [6 x 5 + 3] + [6x 5 + 5] + [6 x 5 + 7]+ [6 x 5 + 9] + [6 x 5 + 11]
=[30 + 1]+ [30 + 3] + [30 + 5] + [30 + 7]+[30+9] + [30+11]
= 31 + 33 + 35 + 37 + 39 + 41 = 216

We have, 1005
One's digit of 1005 = 5
Now, cube of one's digit of 1005 = (5)³ = 5 x 5 x 5 = 125
Hence,one's digit in the cube of 1005 is 5.

30

We have 0.001728 =\(1728\over1000000\)
Now,


\(\therefore {1728\over 1000000}\)
\(={2\times2\times2\times2\times2\times2\times3\times3\times3\over2\times2\times2\times2\times2\times2\times5\times5\times5\times5\times5\times5 }\)
\(={3\times3\times3\over 5\times5\times5\times5\times5\times5}={3^3\over 5^5\times5^3}=({3\over5\times5})^3\)
\(\therefore \sqrt[3]{1728\over 1000000}=({3\over 5\times5})^3\Rightarrow \sqrt[3]{1728\over 1000000}\)
\(={3\over 5\times5}\Rightarrow \sqrt[3]{0.001728}={3\over25}\)
Thus, 0.001728 is the cube of\({3\over25}\) .

0.01

Resolving 9720 into prime factors, we get
9720 =2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 5
The prime factors 3 and 5 are not forming a group of three (triples).
So, 9720 is not a perfect cube.
In prime factorisation of 9720, two factors 3 and 5 remain ungrouped.
So, if we divide the number by 3 x 3 x 5, then the prime factorisation of the quotient will not contain 45.
∴ 9720 + 45 = 216 =(6)3
= 2 x 2 x 2 x 3 x 3 x 3
Hence, the smallest number by which 9720 should be divided to get a perfect cube is 45.

9

13824 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 = 23 x 23 x 23 x 33.
Therefore,  \(\sqrt[3]{13824}=\) 2 x 2 x 2 x 3 = 24