Exercise 4.2
Question 1
Find the cube root of each of the following numbers by prime factorisation:
(i) 12167
(ii) 35937
(iii) 42875
(iv) 21952
(v) 373248
(vi) 32768
(vii) 262144
(viii) 157464
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
Find the cube root of each of the following cube numbers through estimation.
(i) 19683
(ii) 59319
(iii) 85184
(iv) 148877
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
Find the cube root of each of the following numbers:
(i) -250047
(ii) $\dfrac{-64}{1331}$
(iii) $4\dfrac{17}{27}$
(iv) $5\dfrac{1182}{2197}$
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
Find the cube root of each of the following numbers:
(i) $\sqrt[3]{512\times 729}$
(ii) $\sqrt[3]{(-1331)\times (3375)}$
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
(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
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
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
$8^{3}-x^{3} =387$
$512-x^{3}=387$
$x^{3}=512-387=125$
$x^{3}=125=5 \times 5 \times 5=5^{3}$
∴ Therefore the smaller number is 5
Cube of this number is 125
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