243 = 3 x 3 x 3 x 3 x 3
In the above factorisation 3 x 3 remains after grouping the 3’s in triplets. Therefore, 243 is not a perfect cube.

(a) 32768
(b) 2744
(f) 21952

(a) 729
(c) 1728
(e) 6859

By prime factorisation, we have

     

By prime factorisation,
we have
13824 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3
\(\therefore \sqrt[3]{13824}=\)= 2 x 2 x 2 x 3
= 24

Thus, the cube root of 13824 is 24.

We have, 81
Resolving 81 into prime factors, we get
81 = 3 x 3 x 3 x 3
The prime factor 3 does not appear in a group of three.
So, 81 is not a perfect cube.
Thus, we will divide the number by 3.
∴ 81 ÷ 3 =27 = 3 x 3 x 3
Hence, the smallest number by which 81 should be divided to make it a perfect cube is 3.

We have
900 = 2 x 2 x 3 x 3 x 5 x 5
which are ungrouped in triples.

\(\therefore\) 900 is not a perfect cube.

12³ - 11³ =1 + 12 x 11 x 3 =1 +12 x 33 = 1 + 396 = 397

We have, 1024
One's digit of 1024 = 4
Now, cube of one's digit of 1024 = (4)³= 4 x 4 x 4 = 64
Hence,one's digit in the cube of 1024 is 4.

Verification
4104 = 4096 + 8 = 16³ +2³
4104 = 3375 + 729 = 15³ + 9³
13832 = 5832 + 8000 = 18³ + 20³
and 13832 = 13824 + 8 = 24³ + 2³
Hence, it is the required verification.

(1)PERFECT-CUBE
(2)TRIPLES
(3)CUBE-ROOT
(4)ODD NUMBER
(5)PRIME-FACTORS

15

Let 'x' be the side of the cube.
\(\therefore x^3=64 or \sqrt[3]{x^3}=\sqrt[3]{64}\)
\(\Rightarrow \ x =\sqrt[3]{2\times2\times2\times2\times2\times2}=\sqrt[3]{2^3\times2^3}\)
  =2x2=4
Thus, the required side of the cube is 4 cm.

We have

i.e., 1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3
\(\Rightarrow\) 1728 = 2³ x 2³ X 3³
\(\Rightarrow\) 1728 = (2 x 2 x 3)³
\(\Rightarrow \sqrt[3]{1728}\) = 2 x 2 x 3 = 12
Since 1728 is a perfect cube.
\(\therefore\) -1728 is also a perfect cube.
Also\(\sqrt[3]{1728} =-12\)
\(\Rightarrow\) -1728 is a perfect cube of -12.

The prime factorisation of 5832
5832 = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
Therefore, \(\sqrt [ 3 ]{ 5832 } \)
= \(\sqrt [ 3 ]{ \underline { 2\times 2\times 2 } \underline { 3\times 3\times 3 } \underline { 3\times 3\times 3 } } \)
= 2 x 3 x 3
=18

64cm³

11

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\({3\over25}\) .

729 cm³

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

100=2\(\times\)2\(\times\)5\(\times\)5

7\(\times\)7\(\times\)7=343

0\(\times\)0\(\times\)0=0.

10000 = 2 \(\times\) 2 \(\times\) 2\(\times\) 2 \(\times\) 5 \(\times\) 5\(\times\) 5 \(\times\) 5.
= 2⁴ \(\times\) 5⁴ = 2³ \(\times\) 2\(\times\) 5³\(\times\) 5.

8640

(1.2)³ = (1.2) x (1.2) x (1.2) = 1.728