9

4

729 cm³

100

0.01

49

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.

Given, {5² + (12²)^1/2}³
Now, (5)² = 25
(12²)^1/2=(12) ^{2 x 1/2} =(12)¹ =12
So, (25 + 12)³ =(37)³ = 50653

Given, \(\sqrt [ 3 ]{ 27 } +\sqrt [ 3 ]{ 0.008 } +\sqrt [ 3 ]{ 0.064 } \)
Now, \(\sqrt [ 3 ]{ 27 } =\sqrt [ 3 ]{ 3\times 3\times 3 } =3\)
Now \(\sqrt [ 3 ]{ 0.008 } =\sqrt [ 3 ]{ \frac { 8 }{ 1000 } } \)
= \(\sqrt [ 3 ]{ \frac { 2\times 2\times 2 }{ 2\times 2\times 2\times 5\times 5\times 5 } } =\frac { 2 }{ 10 } \)
Now, \(\sqrt [ 3 ]{ 0.064 } =\sqrt [ 3 ]{ \frac { 64 }{ 1000 } } \)
= \(\sqrt [ 3 ]{ \frac { 2\times 2\times 2\times 2\times 2\times 2 }{ 2\times 2\times 2\times 5\times 5\times 5 } } =\frac { 4 }{ 10 } \)
So, \(\sqrt [ 3 ]{ 27 } +\sqrt [ 3 ]{ 0.008 } +\sqrt [ 3 ]{ 0.064 } \)
= \(3+\frac { 2 }{ 10 } +\frac { 4 }{ 10 } \)
LCM of 1 and 10 is 10.
Then, \(\frac { 30+2+4 }{ 10 } =\frac { 36 }{ 10 } \) = 3.6

Yes,we can guess without prime factorisation. Firstly, we separate1331 into two groups (i.e. one's, tens and hundred's place digit and remaining digit).
So, we separate the given number 1331 into two groups as 331 and 1.
Take first group number 331, whose one's digit is 1.
∴ Unit's digit of cube root of 1 = 1
and take second group number 1, whose one's digit is 1.
∴ Unit's digit of cube root of 1= 1
Thus, the cube root of 1331 is 11.
(i) We have, 4913
Here, unit's digit of 4913 = 3
∴ Unit's digit of its cube root = 7
After striking three digits from the right most side of 4913, we get the number 4.
As 1³ = 1 and 2³ = 8
So, 1³ < 4 < 2³
Therefore, the ten's digit of cube root of 4913 is 1.
\(\sqrt [ 3 ]{ 4913 } \)= 17
(ii) We have, 12167
Here, unit's digit of 12167 = 7
:. Unit's digit of its cube = 3
After striking three digits from the rightmost side of
12167, we get the number 12.
As 2³ = 8 and 3³ = 27
So, 2³ < 12< 3³
Therefore, the ten's digit of the cube root of 12167 is 2.
∴   \(\sqrt [ 3 ]{ 12167 } \)=23
(iii) We have, 32768
Unit's digit of 32768 = 8
Unit's digit of its cube root = 2
After striking three digits from the rightmost side of
32768, we get the number 32.
As 3³ = 27 and 4³ = 64
So, 3³ < 32 < 4³
Therefore, the ten's digit of the cube root of 32768 is 3.
∴  \(\sqrt [ 3 ]{ 32768 } \) = 32