ML AGGARWAL CLASS 8 CHAPTER 4 Cubes and Cube Roots Excercise 4.1

 

Exercise 4.1

Question 1

(i) 648

Sol: Expersing it in to prime factors 

$\begin{array}{r|l}2&648\\ \hline 2& 324 \\ \hline 2&162\\ \hline 3& 81\\ \hline 3& 9 \\ \hline 3&27\\ \hline 3& 9 \\ \hline 3&3\\ \hline &1\end{array}$

648 = 2 x 2 x 2 x 3 x 3 x 3 x 3 

= $2^{3} \times 3^{3} \times 3$

Since 3 is left after grouping in triplets 

∴ 648 is not perfect cube 

(ii) 8640 

Expressing it in to prime factors 

  $\begin{array}{r|l}2&8640\\ \hline 2& 4320 \\ \hline 2&2160\\ \hline 2& 1080\\ \hline 2& 540 \\ \hline 2&270\\ \hline2&135 \\ \hline 3&27\\ \hline 3&9\\ \hline 3&3\\ \hline&1\end{array}$

∴ Since 5 is left after grouping in triplets 

8640 is not a perfect cube 

(iii) $729=9 \times 9 \times 9=9^{3}$ is a perfect cube

(iv) $\quad 8000=20 \times 20 \times 20=20^{3}$ is a perfect cuse

Question 2 

(i) 1728

Expressing it into prime factors 

$\begin{array}{r|l}2&1728\\ \hline 2& 864 \\ \hline 2&432\\ \hline 2& 216\\ \hline 2& 108 \\ \hline 2&54\\ \hline 3& 27 \\ \hline 3&9\\ \hline 3&3\end{array}$

$1728=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$

$=2^{3} \times 2^{3} \times 3^{3}$=$=(2 \times 2 \times 3)^{3}$

$=12^{3}$

$\therefore 12^{3}= 1728$ is a perfect cube

And 1728 is the cube of number 12

(ii) 5832

Expressing it into prime factors

$\begin{array}{l|l}2 & 5832 \\\hline 2 & 2916 \\\hline 2 & 1458 \\\hline 3 & 729 \\\hline 3 & 243 \\\hline 3 & 81 \\\hline 3 & 27 \\\hline 3 & 9 \\\hline\end{array}$

$5832=2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$

$\begin{aligned}=2^{3} \times 3^{3} \times 3^{3} &=(2 \times 3 \times 3)^{3} \\ &=18^{3} \end{aligned}$

$\therefore 18^{3}=5832$ is a perfect cube

And 5832 is the cube of number 18.

(iii) 13824  

Expressing it into prime factors

$\begin{array}{c|c}2 & 13824 \\\hline 2 & {6912} \\\hline 2 & 3452 \\\hline 2 & 1728 \\\hline 2 & 864 \\\hline 3 & 432 \\\hline 3 & 144 \\\hline 2 & {78} \\\hline 3 & {16} \\\hline 2 & {8} \\\hline 2 &{4} \\\hline 2 & 2 \\ 2 &  \hline\end{array}$

$13824=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$

=$2^{3} \times 2^{3} \times 2^{3} \times 3^{3}$

$=(2 \times 2 \times 2 \times 3)^{3}$

$13824=24^{3}$ is a perfect cube

(iv) 35937

$\begin{array}{l|l}3 & 35932 \\\hline 3 & 11979 \\\hline 3 & 3993 \\\hline 3 & 1331 \\\hline 3 & 447 \\\hline & 149\end{array}$

$=3 \times 3 \times 3 \times 3 \times 3 \times 149$

∴ It is not a perfect cube 

Question 3

(i) 243 

Expressing it into prime factors

$\begin{array}{l|l}3 & 243 \\\hline 3 & 81 \\\hline 3 & 27 \\\hline 3 & 9 \\\hline 3 & 3 \\\hline\end{array}$

$243=3 \times 3 \times 3 \times 3 \times 3$

If we multiply abore number with 3 

then it becomes = $3 \times 3 \times 3 \times 3 \times 3 \times 3$

=$3^{3} \times 3^{3}=9^{3}=729$, perfect cube

∴ Therefore the smallest number 3 is to be multiplied to 

make the number a perfect cube.

(ii) 3072

Exprssing it into prime factors 

$\begin{array}{l|l}2 & 3072 \\\hline 2 & 1536 \\\hline 2 & 768 \\\hline 2 & 384 \\\hline 2 & 192 \\\hline 2 & 96 \\\hline 2 & 48 \\\hline 2 & 24 \\2 & 12 \\\hline 2 & 6 \\\hline 3 & 3 \\\hline\end{array}$

$3072=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3$

If we multiply the above number with $2 \times 2 \times 3 \times 3$ i.e 36 it will become

$=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 1$

$=2^{3} \times 2 \times 2^{3} \times 2^{3} \times 3^{3}$

$(2 \times 2 \times 2 \times 2 \times 3)^{3}$

$48^{3}=110592$ i.e $3672 \times 36$

∴ Therefore the smallest number 36 is to be multiplied with 3072 to make the number a perfect cube.

(iii) 11979

Expressing it in to prime factors

$\begin{array}{l|l}3 & 11979 \\\hline 3 & 3993 \\\hline 11 & 1331 \\\hline 11 & 121 \\\hline 11 & 11 \\\hline &1\end{array}$

$11979=3 \times 3 \times 11 \times 11 \times 11$

In the above, Prime factors 3 occure twice 11 occures thrice . Therefore the smallest number by which the given number must be multiplied so that the product is a perfect cube i.e 3

Then product = $3 \times 3 \times 3 \times11 \times 11$ 

$=3^{3} \times 11^{3}=33^{2}, 35937=11979 \times 3$

(iv) 19652

Expressing it into prime factors

$\begin{array}{l|l}2 & 19652 \\\hline 2 & 9826 \\\hline 17 & 4913 \\\hline 17 & 289 \\\hline & 17\end{array}$

$19672=2 \times 2 \times 17 \times 17 \times 17$

2 occurs twice , 17 occure thrice therefore the smallest number by which given number must be multiplied So that product is a perfect cube is 2

Then product =$2 \times 2 \times 2 \times 17 \times 17 \times 17$

$2^{3} \times 17^{3}=34^{3}=39,304 = 19652 \times 2$

Question 4

(i) 1536

Expressing it in to prime factors 

$\begin{array}{l|l}2 & 1536 \\\hline 2 & 768 \\\hline 2 & 384 \\\hline 2 & 192 \\\hline 2 & 96 \\\hline 2 & 48 \\2 & 24 \\\hline 2 & {12} \\\hline 2 & 6\end{array}$

$1536=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3$

1536 is not a perfect cube

To make it perfect cube, we should divide the given number by 3, then the prime factorisation of the quotient will not contain 3.

In that case 

        $1536 \div 3=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2=512$

Which is a perfect cube 

So the smallest number by which 1536 must be divided So that quotient is a perfect cube is 3

(ii) 10985

Expressing it into prime factors, we have

$\begin{array}{l|l}5 & 10985 \\\hline 13 & 2197 \\\hline 13 & 169 \\\hline 13 & 13 \\\hline &1\end{array}$

10985= $5 \times 13 \times 13 \times 13$

10985 is not a perfect cube.

To make it perfect cube, we should divide the given number by 5 ,then the prime factorisation of the quotient will not contain 5.

In that case 

 $10985 \div 5=13 \times 13 \times 13=2197$ , Which is a perfect cube 

So, the smallest numbr by which 10985 must be divided 

So that quotient is a perfect cube is 5

(iii) 28672

Expressing it into prime factors

$\begin{array}{l|l}2 & 28672 \\\hline 2 & 14336 \\\hline 2 & 7168 \\\hline 2 & 3584 \\\hline 2 & 1792 \\\hline 2 & 896 \\\hline 2 & 448 \\\hline 2 & 224 \\\hline 2 & 112 \\\hline 2 & 56 \\\hline 2 & 28 \\\hline 2 & 14 \\\hline & 7\end{array}$

$28172=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7$

28672 is not a perfect cube 

To make it a perfect cube, we should divide the given number by 7

Then the prime factorisation will not '7'

In that case $28672 \div 7=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times \times 2$

$=4096$ is a perfect cube

So the smallest number by which given number must be divided 

So that product will become perfect cube is ' 7'

(iv) 13718

Expressing it in the prime factors

$\begin{array}{l|l}2 & 13718 \\\hline 19 & 689 \\\hline 19 & 361 \\\hline 19 & 19 \\\hline & 1 \\\end{array}$

$13718=19 \times 19 \times 19 \times 2 .$

It is not a perfect cube 

To make it a perfecet cube, we should divide the given number by 2 , then prime factorisation will no contain '2'

In that case $13718 \div 2=19 \times 19 \times 19=6855$ is a perfect cube 

So the smallest number 2 must be diivided from given number to make it perfect cube.

Question 5

The volume occupied by one Cuboid is $3 \times 3 \times 5=45$

45 is not a perfect cube 

In order to make it a cube, the number which is to multiplied is $45 \times 3 \times 5 \times 5$ i.e $3 \times 5 \times 5=75$ is to be multiplied in order to make a cube.

So total number of cuboids are needed to form a Cube are 75.

Question 6

Given Surface area of a cubical box is $486 \mathrm{~cm}^{2}$ 

We have , volume of Cubical box is (side) $^{3}$ and

Surface area of Cubical box is $6 \times(\text { side })^{2}$

i.e Let side of a box is 'a' cm

$6 a^{2}=486 \Rightarrow a^{2}=\frac{486}{6}$

$a^{2}=81 = 9 \times 9$

$a=9 \mathrm{~cm}$

Volume of a Cubical box is $a^{3}=9^{3}=729 \mathrm{~cm}^{3}$

Question 7

(i) $125=5 \times 5 \times 5=5^{3}$ , Cube of odd natural number 

$\begin{array}{l|l}5 & 125 \\\hline 5 & 25 \\\hline 5 & 5 \\\hline & 1 \\\end{array}$

(ii) 

 $\begin{array}{l|l}2 & 512\\\hline 2 & 256 \\\hline 2 & 128 \\\hline 2 & 64 \\\hline 2 & 32 \\\hline 2 & 16 \\\hline 2 & 8 \\\hline 2 & 4 \\2 & 2 \\\hline\end{array}$   

$512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

$=2^{3} \times 2^{3} \times 2^{3}=8^{3}$

Cube of even natural number

(iii) $1000=10 \times 10 \times 10=10^{3}$ , Cube of even natural number 

(iv)$2197=13 \times 13 \times 13=13^{3}$, Cube of odd natural number 

(v) 

$\begin{aligned} 4096=4 \times 4 \times 4 \times 4 \times 4 \times 4 &=4^{3} \times 4^{3} \\ &=16^{3}\end{aligned}$ Cube of even natural number 

(vi) $6859=19 \times 19 \times 19=$$19^{3}$ Cube of odd nuatural number.

Question 8

(i) 231 , unit's digit of cube of number is 1

(ii) 358, One's digits of cube of number is 2

(iii) 419 One's digits of cube of number is 9

(iv)725 One 's digits of cube of number is 5

(v)854 One's digits of cube of number is 4

(vi)987 One's digits of cube is 3

(vii)752 One's digits of cube is 8

(viii)893 One's digits of cube is 7.

Question 9

i) $(-13)^{3}=-13\times-13 x-13=(-13)^{3}=-2197$

(ii) $\left(3 \frac{1}{5}\right)^{3}=\left(\frac{16}{5}\right)^{3} \cdot \frac{16 \times 16 \times 16}{5 \times 5 \times 5}=\frac{4096}{125}$

(iii) $\left(-5 \frac{1}{7}\right)^{3}=\left(-\frac{36}{7}\right)^{3} = \frac{-36 x-36 x-36}{7 \times 7 \times 7}= \frac{-46656}{343}$

Exercise 4.2

Question 1

(i) 12167 

Expressing it in to prime factors

$\begin{array}{l|l}23 & 12167 \\\hline 23 & 529 \\\hline &23\end{array}$

$12167=23 \times 23 \times 23$

Hence, Cube root of 12167 is 23

(ii)35937

Expressing it in to prime factors

$\begin{array}{l|l}33 & 35937 \\\hline 33 & 1089\\\hline &33\end{array}$

$35937=33 \times 33 \times 33$

Hence, cube root of 35937 is 33

(iii) 42875

Expressing it in to prime factors

$\begin{array}{l|l}35 & 42875 \\\hline 35 & 1225\\\hline &35\end{array}$

$42875=35 \times 35 \times 35$

Hence, Cube root of 43875is 35

(iv) 21952

Expressing it in to prime factors

$\begin{array}{l|l}2 & 21952 \\\hline 2 & 10976 \\\hline 2 & 5488 \\\hline 2 & 2744 \\\hline 2 & 1372 \\\hline 2 & 686 \\\hline 7 &343 \\\hline 7 & 49 \\\hline&7\end{array}$

$=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7 \times 7 \times 7$

$=(2 \times 2 \times 7)^{3}=(28)^{3}$

Hence, cube root of 21952 is 28

(v) 373248

Expressing it into prime factors

$\begin{array}{l|l}2 & 373248 \\\hline 2 & 186624 \\\hline 2 & 93312 \\\hline 2 & 46656 \\\hline 2 & 23328 \\\hline 2 & 11664 \\\hline 2 & 5832 \\\hline 2 & 2916 \\\hline 2 & 1458 \\\hline 3 &729 \\\hline 3 & 243 \\\hline 3 & 81 \\\hline 3& 27 \\\hline 3&9\\\hline &3\end{array}$

$333248=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$

$=(2 \times 2 \times 2 \times 3 \times 3)^{3}$

$=72^{3}$

Hence, Cube root of 373248 is 72

(vi)32768

Expressing it in to prime factors 

$\begin{array}{l|l}2 & 32768 \\\hline 2 & 16384 \\\hline 2 & 8192 \\\hline 2 & 4096 \\\hline 2 & 2048 \\\hline 2 & 1024 \\\hline 2 & 512 \\\hline 2 & 256 \\\hline 2 & 128 \\\hline 2 &64 \\\hline 2 & 32 \\\hline 2 & 16 \\\hline 2& 4 \\\hline 2&2\\\hline &2\end{array}$

$32768=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

$=(2 \times 2 \times 2 \times 2 \times 2)^{3}$

$=32^{3}$

Hence , Cube root of 32768 is 32.

(vii)262144

Expressing it in to prime factors 

$\begin{array}{l|l}2 & 262144 \\\hline 2 & 131072 \\\hline 2 & 65536 \\\hline 2 & 32768 \\\hline 2 & 16384 \\\hline 2 & 8192 \\\hline 2 & 4096 \\\hline 2 & 2048 \\\hline 2 & 1024 \\\hline 2 &512 \\\hline 2 & 256 \\\hline 2 & 128 \\\hline 2& 64 \\\hline 2&32\\\hline 2&16 \\\hline 2& 8\\\hline 2&4\\\hline 2&2\\\hline &1\end{array}$

$262144=\underbrace{2 \times 2 \times 2} \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

$=(2 \times 2 \times 2 \times 2 \times 2 \times 2)^{3}$

$=64^{3}$

Hence, Cube root of 262144 is 64

(viii) 157464

Expressing it into prime factors

$\begin{array}{l|l}2 & 157464 \\\hline 2 & 78732 \\\hline 2 & 39366 \\\hline 3 & 19683 \\\hline 3 & 6561 \\\hline 3 & 2187 \\\hline  & 48 \\\hline 2 & 24 \\2 & 12 \\\hline 2 & 6 \\\hline 3 & 3 \\\hline\end{array}$

$=2^{3} \times 3^{3} \times 3^{3} \times 3^{3}$

$=(2 \times 3 \times 3 \times 3)^{3}$

$=54^{3}$

Hence , Cube root of 157464 is 54

Question 2 

(i) 19683

$\frac{19}{\text { second group }}$ 

$\frac{683}{\text { First group }}$

First gruop decides the unit digits of required cube root 

So, The number 683 ends with 3 . we know that 3 comes 

at units place of a number only when it's when it's cube root 

end in 7 

Now take second group 19 , then it will decide the ten's digit of required cube root

Now that $2^{3}=8$ and $3^{3} = 27$ ,Also $8<19<27 .$

We take the one's place of smaller number 8 as ten's digit of required cube root (i.e 2)

Therefore $\sqrt[3]{19683}=27$ 

(ii) 59319 

$\frac{59}{\text { second group }}$

$\frac{319}{\text { first group }}$

First group decides the one's digits of required Cube root 

The number 319 ends with 9. we know that 9 comes at 

Unit's place of a number only when its cube root ends in 9 

Now second group decides the ten's digits of required cube root 

59 lies in between $3^{3}=27$ and $4^{3}=64$. We take one's 

place of smaller number 27 as the ten's digits of required cube 

root

So $\sqrt[3]{59319}=27$

(iii) 85184

$\frac{85}{\text {second group }}$

$\frac{184}{\text { First group }}$

184 ends with 4 . we know that 4 comes at unit's 

place of number only when its cube root ends in 4

Second group decides ten's digits 

i.e 85 lies in between $4^{3}=64$ and $5^{3}=125$

we know that one's place of smaller number 64 as 

ten's digits of required Cube root 

So $\sqrt[3]{85184}=44$

(iv) 148877

$\frac{148}{\text { second group }}$

$\frac{877}{\text { First group }}$

Step:1 First form group of three digits starting from rightmost 

digit (i.e unit's digits ) of number 

Step:2 First group decides unit's digits of required root 

The number 877 ends with 7. we know that 7 comes at 

unit's place of a number only when it's cube root ends in

3

So the unit digit of required Cube root is 3

Step:3 If no group is left then number obtained is the cube root of given number 

But if second group exits (in this case 148) then it will decide the ten's digit of required 

cube root 

Now take second group i.e 148

We know that $5^{3}=125$ and $6^{3}=216$ . Also $125<148<216$ 

We take one's place of the smaller number 725 as the ten's digit of required 

cube root (i.e 5)

Step: 4 If no group is left then the digit obtained in step 2 and 

step 3 decides the cube root of given number 

i.e $\sqrt[3]{148877}=53$.

Question 3

(i) 250047

Expressing it into to prime factors 

-250047= $7\times -7\times -7 \times -3 \times -3 \times-3\times -3\times -3\times -3$ 

= $(-7\times -3\times -3)^{3}$

$=(-63)^{3}$

Hence , Cube root of -250047 is -63

(ii) $\frac{-64}{1331}$

Expressing 64 and 1331 in to prime factors 

$\begin{aligned} 64 &=4 \times 4 \times 4=4^{3} \\ 1331 &=11 \times 11 \times 11=11^{3} \end{aligned}$

$\frac{-64}{1331}=\frac{(-4)^{3}}{(11)^{3}}=\left(\frac{-4}{11}\right)^{3} \Rightarrow \sqrt[3]{\frac{-6 y}{1331}}=\frac{-4}{11}$

(iii) $4 \frac{17}{27}=\frac{125}{27}$

Expresssing 125 and 27 it into prime factors 

$125= 5 \times 5 \times 5=5^{3}$

$27=3 \times 3 \times 3=3^{3}$

Hence,$\sqrt[3]{\frac{125}{27}}=\sqrt[3]{\left(\frac{5}{3}\right)^{3}}=\frac{5}{3} .$

(iv)$5 \frac{1182}{2197}=\frac{12167}{2197}$

$12167=23 \times 23 \times 23=23^{3}$

$2193=13 \times 13 \times 13=13^{3}$

$\frac{12167}{2197}=\frac{23^{3}}{13^{3}} =\left(\frac{23}{13}\right)^{3}$

Hence, Cube root of $5 \frac{1182}{2197}$ is $\frac{23}{13}$

Question 4

(i) $\sqrt[3]{512 \times 729}$

Expressing it into prime factors 

 $512=8 \times 8 \times 8=8^{3}$

$729=9 \times 9 \times 9=9^{3}$

$512 \times 729=8^{3} \times 9^{3}$= $(8 \times 9)^{3}=72^{3}$

Hence, $\sqrt[3]{512 \times 729}$ = $\sqrt[3]{72^{3}}$= 72

(ii) $\sqrt[3]{(-1331) \times(3335)}$

Expressing it into prime factors 

$\begin{array}{l|l}11 & 1131 \\\hline 11 & 121\\\hline11&11\\\hline &1\end{array}$

$\begin{array}{l|l}5 & 3375\\\hline 5 & 675 \\\hline 5 & 135 \\\hline 3 & 27 \\\hline 3 & 9 \\\hline 3 & 3\\\hline&1\end{array}$

$-1331=(-11)^{3}$

$\begin{aligned} 3375 \times 5 \times 5 \times 3 \times 3 \times 3=&(5 \times 3)^{3} \\ &=15^{3} \end{aligned}$ 

$-1331 \times 3375=(-11 \times 15)^{3}$

Hence $\sqrt[3]{(-1331 \times 3] 75)}=\sqrt[3]{(-11 \times 15)^{3}}$ = $-11 \times 15$

$=-165$

Question 5

(i) 0.003375

$\sqrt[3]{0.003335}=\sqrt[3]{\frac{3375}{1000000}}$

$=\sqrt[3]{\frac{15 \times 15 \times 15}{100 \times 100 \times 100}}$

$=\frac{15}{100}=0.15$

(ii) 19.683

$\sqrt[3]{19.683}=\sqrt[3]{\frac{19683}{1000}}$

$=\sqrt[3]{\frac{27 \times 27 \times 27}{6 \times 10 \times 10}}$\

$=\frac{27}{10}=2.7$

$\begin{array}{l|l}3 & 19683 \\\hline 3& 6561 \\\hline 3 & 2187 \\\hline 3 & 729 \\\hline 3 & 243 \\\hline 3 & 81 \\\hline 3 &27 \\\hline 3 & 9 \\\hline&3\end{array}$

=$3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$

=$(3 \times 3 \times 3)=27^{3}$

Question 6

$\sqrt[3]{27}+\sqrt[3]{0.008}+\sqrt[3]{0.664}$

$27=3 \times 3 \times 3 \Rightarrow \sqrt[3]{27}: \sqrt[3]{3 \times 3 \times 3}=3$

$\sqrt[3]{0 \cdot 008}=\sqrt[3]{\frac{8}{1000}}=\sqrt[3]{\frac{2 \times 2 \times 2}{10 \times 10 \times 10}}=\frac{2}{10}=0.2$

$\sqrt[3]{0.064}=\sqrt[3]{\frac{64}{1000}}=\sqrt[3]{\frac{4 \times 4 \times 4}{10 \times 10 \times 10}}=\frac{4}{10}=0.4$

$\sqrt[3]{27}+\sqrt[3]{0.008}+\sqrt[3]{0.064}, 3+0.2+0.4=3.6$

Question 7

Sol: 6561

Expressing it into prime factors

$\begin{array}{l|l}3 & 6561 \\\hline 3& 2187 \\\hline 3 & 729 \\\hline 3 & 243 \\\hline 3 &81 \\\hline 3 & 27 \\\hline 3 &9 \\\hline 3 & 3 \\\hline&1\end{array}$

$6561=3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$

$=27 \times 27 \times 9$

9 is left in the above expansion, if we multiply 

the above number with 3 i.e $9\times 3$ = 27

i.e it becomes 

$6561 \times 3=27 \times 27 \times 9 \times 3=27 \times 27 \times 27=19683-27^{3}$

So the smallest number 3 must be multiplied to become 

the number a perfect cube 

Cube root of 19683 = 27

Question 8

Sol: 8748

Expressing it in to prime factors

$8748=3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 2 \times 2$

$=27 \times 27 \times 3 \times 2 \times 2$

If we divide above numbr by $\frac{4}{9}$

then it becomes $=\frac{27 \times 27 \times 3 \times 4}{(4 / 9)}$

$27 \times 27 \times 27$ =$27^{3}$ 

$=19683$

$\frac{8748 \times 9}{4}=19683=27^{3}$

Question 9

Givne Volume of cubical box = $21952 \mathrm{~m}^{3}$

We know that Volume of cube = (Side) $^{3}$

Let side length of cube = a

$a^{3}=21952$ 

$\begin{array}{l|l}4 & 21952 \\\hline 4& 5488 \\\hline 4 &1372 \\\hline 7 &343 \\\hline 7 &49 \\\hline \\\hline&7\end{array}$

$=4 \times 4 \times 4 \times 7 \times 7 \times 7$

$a^{3}=(4 \times 7)^{3}$

$a=28 m$

∴ Length of side of box = 28 m

Question 10

Let the three number be $3 x, 4 x, 5 x$, then

$(3 x) \times(4 x)(5 x)=480$

$60 x^{3}=480$

$x^{3}=\frac{480}{60}=8$

$x^{3}=8=2^{3}$

∴ $x=2 \Rightarrow 3 x=6, \quad 4 x=8,5 x=10$

Question 11

Let the two numbers are $4 x, 5 x$, Then

$(5 x)^{3}-(4 x)^{3}=61$

$125 x^{3}-64 x^{3}=61$

$61 x^{3}=61$

$x^{3}=\frac{61}{61}=1$

$x^{3}=1 \Rightarrow x=1$

∴ The number are 4, 5

Question 12

Let the cube root of smaller number be x

Given 

$8^{3}-x^{3} =387$

$512-x^{3}=387$

$x^{3}=512-387=125$

$x^{3}=125=5 \times 5 \times 5=5^{3}$

$x^{3}=5^{3}$

$x=5$

∴ Therefore the smaller number is 5 

Cube of this number is 125