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

11

10

42

5

A cube has 6 equal surfaces.
Let side of the cube be x.
\(\therefore\)Area of a surface =x²
\(\Rightarrow\) Area of the given cube = 6x²
\(\therefore\)6x²= 486 \(\Rightarrow x^2={486\over6}=81\)
\(\Rightarrow x =\sqrt{81}=9\)
Thus, the volume of the given cube = 729 cm³ .

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., Resolving 216 into prime factors we have:
216 = 2 x 2 x 2 x 3 x 3 x 3
i.e., 216 can be resolved in to such prime factor which can be grouped into triple of equal of factor and no factor is left over.
\(\therefore\)216 is a perfect cube.
Now, 216 = 2³ x 3³ \(\Rightarrow\) 216 = (2 x 3)³ = 6³
\(\Rightarrow \sqrt[3]{216 }\) = \(\sqrt[3]{6^3}\)\(\Rightarrow \sqrt[3]{216 }\) = 6
Thus, the required number is 6, whose cube is 216.

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.

We have 31944 = 2 x 2 x 2 x 3
x11x11x11
Since, the prime factors of 31944 do not appear in triples as 3 is left over.
\(\therefore\)31944 is not a perfect cube.
Obviously, 31944 \(\div\) 3 will be a perfect cube
i.e., [31944] \(\div\) 3
= [2 x 2 x 2 x 3 x 11 x 11 x11] \(\div\) 3
or 10648 = 2 x 2 x 2 x 11 x 11 x 11
\(\therefore\) 10648 is a perfect cube.
Thus, the required least number = 3.

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.
\(\therefore\) 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.
 

\(\frac { 2197 }{ 125 } \)

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

Given, number is 614125.
So, groups of 614125 are

In the first group, the number 125 ends with 5. We know that, 5 comes at the unit's place of a number only when its cube ends in 5.
So, 5 will come at unit's place.
Now, in the second group, the number is 614
∵ (8)³ = 512 and (9)³=729
and 512 < 614 < 729
So, ten's place of required cube root is 8.
Hence, \(\sqrt [ 3 ]{ 614125 } \) = 85

Given, number is 68600.
Now, prime factorisation of 68600
68600 = 2 x 2 x 2 x 5 x 5 x 7 x 7 x 7
The prime factor 5 does not have triples.
∴ 68600 is not a perfect cube
So, if we multiply the number 68600 by 5, then we get a perfect cube.
i.e., 68600 x 5 = 343000
= 2 x 2 x 2 x 5 x 5 x 5 x 7 x 7 x 7
= 2 x 5 x 7 =70