Class 8 Math Topic 3 - Indices And Cube Root

Question 1:

Express the following numbers in index form.
(1) Fifth root of 13
(2) Sixth root of 9
(3) Square root of 256
(4) Cube root of 17
(5) Eighth root of 100
(6) Seventh root of 30

ANSWER:

It is known that,

nth root of a is expressed as a1/n.

(1) Fifth root of 13

= (13)1/5

(2) Sixth root of 9

= (9)1/6

(3) Square root of 256

= (256)1/2

(4) Cube root of 17

= (17)1/3

(5) Eighth root of 100

= (100)1/8

(6) Seventh root of 30

= (30)1/7

Question 2:

Write in the form 'nth root of a' in each of the following numbers.
(1) (81)14${\left(81\right)}^{\frac{1}{4}}$

(2) 4912${49}^{\frac{1}{2}}$

(3) (15)15${\left(15\right)}^{\frac{1}{5}}$

(4) (512)19${\left(512\right)}^{\frac{1}{9}}$

(5) 100119${100}^{\frac{1}{19}}$

(6) (6)17${\left(6\right)}^{\frac{1}{7}}$

ANSWER:

It is known that,

nth root of a is expressed as a1/n.
(1) (81)14$\left(1\right) {\left(81\right)}^{\frac{1}{4}}$
= Fourth root of 81
(2) (49)12$\left(2\right) {\left(49\right)}^{\frac{1}{2}}$
= Square root of 49
(3) (15)15$\left(3\right) {\left(15\right)}^{\frac{1}{5}}$
= Fifth root of 15
(4) (512)19$\left(4\right) {\left(512\right)}^{\frac{1}{9}}$
= Ninth root of 512
(5) (100)119$\left(5\right) {\left(100\right)}^{\frac{1}{19}}$
= Nineteenth root of 100
(6) (6)17$\left(6\right) {\left(6\right)}^{\frac{1}{7}}$
= Seventh root of 6

Question 1:

Complete the following table.

Sr. No.	Number	Power of the root	Root of the power
(1)	(225)32${\left(225\right)}^{\frac{3}{2}}$	Cube of square root of 225	Square root of cube of 225
(2)	(45)45${\left(45\right)}^{\frac{4}{5}}$
(3)	(81)67${\left(81\right)}^{\frac{6}{7}}$
(4)	(100)410${\left(100\right)}^{\frac{4}{10}}$
(5)	(21)37${\left(21\right)}^{\frac{3}{7}}$

ANSWER:

Sr. No.	Number	Power of the root	Root of the power
(1)	(225)32${\left(225\right)}^{\frac{3}{2}}$	cube of square root of 225	square root of cube of 225
(2)	(45)45${\left(45\right)}^{\frac{4}{5}}$	4th power of 5th root of 45	5th root of 4th power of 45
(3)	(81)67${\left(81\right)}^{\frac{6}{7}}$	6th power of 7th root of 81	7th root of 6th power of 81
(4)	(100)410${\left(100\right)}^{\frac{4}{10}}$	4th power of 10th root of 100	10th root of 4th power of 100
(5)	(21)37${\left(21\right)}^{\frac{3}{7}}$	cube of 7th root of 21	7th root of cube of 21

Question 2:

Write the following numbers in the form of rational indices.
(1) Square root of 5th power of 121
(2) Cube of 4th root of 324
(3) 5th root of square of 264
(4) Cube of cube root of 3

ANSWER:

It is known that,

am/n = (am)1/n means 'nth root of mth power of a'.

am/n = (a1/n)m means 'mth power of nth root of a'.

(1) Square root of 5th power of 121
={(121)5}12=(121)52$$\begin{gathered} ={\left\{{\left(121\right)}^5\right\}}^{\frac{1}{2}} \\ ={\left(121\right)}^{\frac{5}{2}} \end{gathered}$$
(2) Cube of 4th root of 324
={(324)14}3=(324)34$$\begin{gathered} ={\left\{{\left(324\right)}^{\frac{1}{4}}\right\}}^3 \\ ={\left(324\right)}^{\frac{3}{4}} \end{gathered}$$
(3) 5th root of square of 264
={(264)2}15=(264)25$$\begin{gathered} ={\left\{{\left(264\right)}^2\right\}}^{\frac{1}{5}} \\ ={\left(264\right)}^{\frac{2}{5}} \end{gathered}$$
(4) Cube of cube root of 3
={313}3=333$$\begin{gathered} ={\left\{3^{\frac{1}{3}}\right\}}^3 \\ =3^{\frac{3}{3}} \end{gathered}$$

Question 1:

Find the cube roots of the following numbers.
(1) 8000
(2) 729
(3) 343
(4) −512
(5) −2744
(6) 32768

ANSWER:

(1) To find the cube root of 8000, let us factorise 8000 first.
8000=2×2×2×2×2×2×5×5×58000=4×4×4×5×5×5 = 43×53=(4×5)3    [am×bm=(a×b)m]8000=(20)3∴8000−−−−√3=20$$\begin{gathered} 8000=2\times 2\times 2\times 2\times 2\times 2\times 5\times 5\times 5 \\ 8000=4\times 4\times 4\times 5\times 5\times 5 = 4^3\times 5^3={\left(4\times 5\right)}^3 \left[a^m\times b^m={\left(a\times b\right)}^m\right] \\ 8000={\left(20\right)}^3 \\ \therefore \sqrt[3]{8000}=20 \end{gathered}$$
(2) To find the cube root of 729, let us factorise 729 first.
729=3×3×3×3×3×3729=(3×3)×(3×3)×(3×3)=(3×3)3=93∴729−−−√3=9$$\begin{gathered} 729=3\times 3\times 3\times 3\times 3\times 3 \\ 729=\left(3\times 3\right)\times \left(3\times 3\right)\times \left(3\times 3\right)={\left(3\times 3\right)}^3=9^3 \\ \therefore \sqrt[3]{729}=9 \end{gathered}$$
(3) To find the cube root of 343, let us factorise 343 first.
343=7×7×7=73∴343−−−√3=7$$\begin{gathered} 343=7\times 7\times 7=7^3 \\ \therefore \sqrt[3]{343}=7 \end{gathered}$$
(4) To find the cube root of  −512, let us factorise 512 first.
512=2×2×2×2×2×2×2×2×2512=(2×2×2)×(2×2×2)×(2×2×2)=8×8×8×=83−512=(−8)×(−8)×(−8)=(−8)3∴−512−−−−√3=−8$$\begin{gathered} 512=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2 \\ 512=\left(2\times 2\times 2\right)\times \left(2\times 2\times 2\right)\times \left(2\times 2\times 2\right)=8\times 8\times 8\times =8^3 \\ -512=\left(-8\right)\times \left(-8\right)\times \left(-8\right)={\left(-8\right)}^3 \\ \therefore \sqrt[3]{-512}=-8 \end{gathered}$$
(5) To find the cube root of −2744, let us factorise 2744 first.
2744=2×2×2×7×7×72744=(2×7)×(2×7)×(2×7)=(2×7)3=143−2744=(−14)3∴−2744−−−−−√3=−14$$\begin{gathered} 2744=2\times 2\times 2\times 7\times 7\times 7 \\ 2744=\left(2\times 7\right)\times \left(2\times 7\right)\times \left(2\times 7\right)={\left(2\times 7\right)}^3={14}^3 \\ -2744={\left(-14\right)}^3 \\ \therefore \sqrt[3]{-2744}=-14 \end{gathered}$$
(6) To find the cube root of 32768, let us factorise 32768 first.
32768=2×2×2×2×2×2×2×2×2×2×2×2×2×2×232768=(2×2×2×2×2)×(2×2×2×2×2)×(2×2×2×2×2)=32×32×32=(32)3∴32768−−−−−√3=32$$\begin{gathered} 32768=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2 \\ 32768=\left(2\times 2\times 2\times 2\times 2\right)\times \left(2\times 2\times 2\times 2\times 2\right)\times \left(2\times 2\times 2\times 2\times 2\right)=32\times 32\times 32={\left(32\right)}^3 \\ \therefore \sqrt[3]{32768}=32 \end{gathered}$$

Question 2:

Simplify:
(1) 27125−−−√3$\sqrt[3]{\frac{27}{125}}$

(2) 1654−−√3$\sqrt[3]{\frac{16}{54}}$

ANSWER:

(1) 27125−−−√3=3×3×35×5×5−−−−−√3=3353−−√3=(35)3−−−−√3     [ambm=(ab)m]=35$$\begin{gathered} \left(1\right) \sqrt[3]{\frac{27}{125}} \\ =\sqrt[3]{\frac{3\times 3\times 3}{5\times 5\times 5}} \\ =\sqrt[3]{\frac{3^3}{5^3}} \\ =\sqrt[3]{{\left(\frac{3}{5}\right)}^3} \left[\frac{a^m}{b^m}={\left(\frac{a}{b}\right)}^m\right] \\ =\frac{3}{5} \end{gathered}$$
(2) 1654−−√3=2×2×2×22×3×3×3−−−−−−−√3=2×2×23×3×3−−−−−√3=2333−−√3=(23)3−−−−√3    [ambm=(ab)m]=23$$\begin{gathered} \left(2\right) \sqrt[3]{\frac{16}{54}} \\ =\sqrt[3]{\frac{2\times 2\times 2\times 2}{2\times 3\times 3\times 3}} \\ =\sqrt[3]{\frac{2\times 2\times 2}{3\times 3\times 3}} \\ =\sqrt[3]{\frac{2^3}{3^3}} \\ =\sqrt[3]{{\left(\frac{2}{3}\right)}^3} \left[\frac{a^m}{b^m}={\left(\frac{a}{b}\right)}^m\right] \\ =\frac{2}{3} \end{gathered}$$

Question 3:

If 729−−−√3$\sqrt[3]{729}$ = 9 then 0.000729−−−−−−−√3$\sqrt[3]{0.000729}$ = ?

ANSWER:

It is given that,
729−−−√3=90.000729−−−−−−−√3=7291000000−−−−−√3=729√31000000√3=93√3(100)3√3=9100=0.09