Mathematics Solutions for Class 8 Maths Chapter 3 - Indices And Cube Root

Answer:

It is known that,

n^th root of a is expressed as a^1/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

Answer:

It is known that,

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

Answer:

Answer:

It is known that,

a^m/n = (a^m)^1/n means 'n^th root of m^th power of a'.

a^m/n = (a^1/n)^m means 'm^th power of n^th root of a'.

(1) Square root of 5^th power of 121
$$\begin{gathered} ={\left\{{\left(121\right)}^5\right\}}^{\frac{1}{2}} \\ ={\left(121\right)}^{\frac{5}{2}} \end{gathered}$$
(2) Cube of 4^th root of 324
$$\begin{gathered} ={\left\{{\left(324\right)}^{\frac{1}{4}}\right\}}^3 \\ ={\left(324\right)}^{\frac{3}{4}} \end{gathered}$$
(3) 5^th root of square of 264
$$\begin{gathered} ={\left\{{\left(264\right)}^2\right\}}^{\frac{1}{5}} \\ ={\left(264\right)}^{\frac{2}{5}} \end{gathered}$$
(4) Cube of cube root of 3
$$\begin{gathered} ={\left\{3^{\frac{1}{3}}\right\}}^3 \\ =3^{\frac{3}{3}} \end{gathered}$$

Answer:

(1) To find the cube root of 8000, let us factorise 8000 first.
$$\begin{gathered} 8000=2\times 2\times 2\times 2\times 2\times 2\times 5\times 5\times 5 \\ 8000=4\times 4\times 4\times 5\times 5\times 5 = 4^3\times 5^3={\left(4\times 5\right)}^3 \left[a^m\times b^m={\left(a\times b\right)}^m\right] \\ 8000={\left(20\right)}^3 \\ \therefore \sqrt[3]{8000}=20 \end{gathered}$$
(2) To find the cube root of 729, let us factorise 729 first.
$$\begin{gathered} 729=3\times 3\times 3\times 3\times 3\times 3 \\ 729=\left(3\times 3\right)\times \left(3\times 3\right)\times \left(3\times 3\right)={\left(3\times 3\right)}^3=9^3 \\ \therefore \sqrt[3]{729}=9 \end{gathered}$$
(3) To find the cube root of 343, let us factorise 343 first.
$$\begin{gathered} 343=7\times 7\times 7=7^3 \\ \therefore \sqrt[3]{343}=7 \end{gathered}$$
(4) To find the cube root of  −512, let us factorise 512 first.
$$\begin{gathered} 512=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2 \\ 512=\left(2\times 2\times 2\right)\times \left(2\times 2\times 2\right)\times \left(2\times 2\times 2\right)=8\times 8\times 8\times =8^3 \\ -512=\left(-8\right)\times \left(-8\right)\times \left(-8\right)={\left(-8\right)}^3 \\ \therefore \sqrt[3]{-512}=-8 \end{gathered}$$
(5) To find the cube root of −2744, let us factorise 2744 first.
$$\begin{gathered} 2744=2\times 2\times 2\times 7\times 7\times 7 \\ 2744=\left(2\times 7\right)\times \left(2\times 7\right)\times \left(2\times 7\right)={\left(2\times 7\right)}^3={14}^3 \\ -2744={\left(-14\right)}^3 \\ \therefore \sqrt[3]{-2744}=-14 \end{gathered}$$
(6) To find the cube root of 32768, let us factorise 32768 first.
$$\begin{gathered} 32768=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2 \\ 32768=\left(2\times 2\times 2\times 2\times 2\right)\times \left(2\times 2\times 2\times 2\times 2\right)\times \left(2\times 2\times 2\times 2\times 2\right)=32\times 32\times 32={\left(32\right)}^3 \\ \therefore \sqrt[3]{32768}=32 \end{gathered}$$

Answer:

$$\begin{gathered} \left(1\right) \sqrt[3]{\frac{27}{125}} \\ =\sqrt[3]{\frac{3\times 3\times 3}{5\times 5\times 5}} \\ =\sqrt[3]{\frac{3^3}{5^3}} \\ =\sqrt[3]{{\left(\frac{3}{5}\right)}^3} \left[\frac{a^m}{b^m}={\left(\frac{a}{b}\right)}^m\right] \\ =\frac{3}{5} \end{gathered}$$
$$\begin{gathered} \left(2\right) \sqrt[3]{\frac{16}{54}} \\ =\sqrt[3]{\frac{2\times 2\times 2\times 2}{2\times 3\times 3\times 3}} \\ =\sqrt[3]{\frac{2\times 2\times 2}{3\times 3\times 3}} \\ =\sqrt[3]{\frac{2^3}{3^3}} \\ =\sqrt[3]{{\left(\frac{2}{3}\right)}^3} \left[\frac{a^m}{b^m}={\left(\frac{a}{b}\right)}^m\right] \\ =\frac{2}{3} \end{gathered}$$

Answer:

It is given that,
$$\begin{gathered} \sqrt[3]{729}=9 \\ \sqrt[3]{0.000729}=\sqrt[3]{\frac{729}{1000000}} \\ =\frac{\sqrt[3]{729}}{\sqrt[3]{1000000}} \\ =\frac{\sqrt[3]{9^3}}{\sqrt[3]{{\left(100\right)}^3}} \\ =\frac{9}{100} \\ =0.09 \end{gathered}$$