ML AGGARWAL CLASS 8 CHAPTER 4 Cubes and Cube Roots Exercise 4.2

 Exercise 4.2

Question 1

Sol :
(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 

Sol :
(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

Sol :

(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) $4 \frac{17}{27}=\frac{125}{27}$

Expressing 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

Sol :

(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

Sol :

(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

Sol :

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

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

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

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

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


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

Sol :
Given : 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

Sol :

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

Sol :

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

Sol :

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

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