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

Exercise 4.1

Question 1

(i) 648

Sol: Expresing 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 

Sol :

(i) 1728

Sol :

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

Sol :

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  

Sol :

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

Sol :

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

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

∴ It is not a perfect cube 

Question 3

Sol :

(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

Expressing 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

Sol :

(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

Sol :

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

Sol :

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

Sol :

(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

Sol :

(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

Sol :

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