Exercise 7.2 Cubes and Cube Roots- NCERT Solutions Class 8

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Question 1

Find the cube root of each of the following numbers by prime factorization method.

(i) \(64\)

(ii) \(512\)

(iii) \(10648\)

(iv) \(27000\)

(v) \(15625\)

(vi) \(13824\)

(vii) \(110592\)

(viii) \(46656\)

(ix) \(175616\)

(x) \(91125\)

Solution

Video Solution

Reasoning:

Factors in the prime factorization of cube should be grouped as triplets.

Steps:

(i)

\(\begin{align} 64&=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2} \\ & ={{2}^{3}}\times {{2}^{3}} \\  \sqrt[3]{64}&=2\times 2=4  \end{align}\)

(ii) 

\(\begin{align}512 =&\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\\&\times \underline{2\times 2\times 2} \\ & ={{2}^{3}}\times {{2}^{3}}\times {{2}^{3}} \\ \sqrt[3]{512}&=2\times 2\times 2=8 \end{align}\)

(iii)

\(\begin{align}10648&=\underline{2\times 2\times 2}\times \underline{11\times 11\times 11} \\  & ={{2}^{3}}\times {{11}^{3}} \\  \sqrt[3]{10648}&=2\times 11=22  \end{align}\)

(iv) 

\(\begin{align} 27000&=\begin{Bmatrix}\underline{2\times 2\times 2}\times \underline{3\times 3\times 3}\\\times \underline{5\times 5\times 5}\end{Bmatrix} \\ & ={{2}^{3}}\times {{3}^{3}}\times {{5}^{3}} \\ \sqrt[3]{27000}&=2\times 3\times 5=30 \end{align}\)

(v) 

\(\begin{align}15625&=\underline{5\times 5\times 5}\times \underline{5\times 5\times 5} \\  & ={{5}^{3}}\times {{5}^{3}} \\ \sqrt[3]{15625}&=5\times 5=25  \end{align}\)

(vi) 

\(\begin{align}13824&=\begin{Bmatrix}\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\\\times \underline{2\times 2\times 2}\times \underline{3\times 3\times 3}\end{Bmatrix} \\ & ={{2}^{3}}\times {{2}^{3}}\times {{2}^{3}}\times {{3}^{3}} \\ \sqrt[3]{13824}&=2\times 2\times 2\times 3=24 \end{align}\)

(vii)

\(\begin{align}110592&= \begin{Bmatrix} \underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\\\times \underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\\\times \underline{3\times 3\times 3}\end{Bmatrix} \\ & ={{2}^{3}}\times {{2}^{3}}\times {{2}^{3}}\times {{2}^{3}}\times {{3}^{3}} \\ \sqrt[3]{110592}&=2\times 2\times 2\times 2\times 3=48 \end{align}\)

(viii)

\(\begin{align}46656& =\begin{Bmatrix}\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\\\times \underline{3\times 3\times 3}\times \underline{3\times 3\times 3} \end{Bmatrix}\\ & ={{2}^{3}}\times {{2}^{3}}\times {{2}^{3}}\times {{3}^{3}}\times {{3}^{3}} \\ \sqrt[3]{46656} & =2\times 2\times 3\times 3=36 \end{align}\)

(ix)

\(\begin{align}175616&=\begin{Bmatrix}\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\\\times \underline{2\times 2\times 2}\times \underline{7\times 7\times 7} \end{Bmatrix}\\ & ={{2}^{3}}\times {{2}^{3}}\times {{2}^{3}}\times {{7}^{3}} \\ \sqrt[3]{175616} &=2\times 2\times 2\times 7=56 \end{align}\)

(x)

\(\begin{align}91125&=\begin{Bmatrix}\underline{5\times 5\times 5}\times \underline{3\times 3\times 3}\\\times \underline{3\times 3\times 3}\end{Bmatrix} \\ & ={{5}^{3}}\times {{3}^{3}}\times {{3}^{3}} \\ \sqrt[3]{91125}& =5\times 3\times 3=45 \end{align}\)

Question 2

 State True or False:

(i). Cube of any odd number is even.

(ii). A perfect cube does not end with two zeros.

(iii). If square of a number ends with \(5,\) then its cube ends with \(25\).

(iv). There is no perfect cube which ends with \(8\).

(v). The cube of a \(2\)-digit number may be a \(3\)-digit number.

(vi). The cube of a \(2\)-digit number may have seven or more digits.

(vii).The cube of a single digit number may be a single digit number.

Solution

Video Solution

(i). Cube of any odd number is even: 

Ans.  False

Reasoning:

Cubes of odd numbers are odd.

Cubes of even numbers are even.

(ii). A perfect cube does not end with two zeros:

Ans.  True

Reasoning:

Perfect cube may end with \(3\) zeros (or) groups of \(3\) zeros.

(iii). If square of a number ends with \(5,\) then its cube ends with \(25\)

Ans.  False

Reasoning:

It is not always necessary that if the square of a number ends with \(5\), then its cube will end with \(25\).

For example, the square of \(5\) is \(25\) and \(25\) has its unit digit as \(5\). The cube of \(5\) is \(125\).

However, the square of \(15\) is \(225\) and also has its unit place digit as \(5\) but the cube of \(15\) is \(3375\) which does not end with \(25\).

(iv). There is no perfect cube which ends with \(8\)

Ans.  False

Reasoning:

The cubes of all the numbers having their unit place digit as \(2\) will end with \(8\).

The cube of \(12\) is \(1728\) and the cube of \(22\) is \(10648\).

(v). The cube of a \(2\)-digit number may be a \(3\)-digit number: 

Ans.  False

Reasoning:

Cube of a \(1\)-digit number may have \(1\)-digit to \(3\)-digits.

Cube of a \(2\)-digit number may have \(4\)-digits to maximum \(6\)-digits

(vi). The cube of a \(2\)-digit number may have seven or more digits: 

Ans.  False

Reasoning:

Cube of a \(1\)-digit number may have \(1\)-digit to \(3\)-digits.

Cube of a \(2\)-digit number may have \(4\)-digits to maximum \(6\)-digits.

(vii).The cube of a single digit number may be a single digit number: 

Ans.  True

Reasoning:

Some examples

\[\begin{align}{1^3}&= 1\\{2^3}& = 8\end{align}\]

Question 3

You are told that \(1,331\) is a perfect cube. Can you guess without factorization what its cube root is?

Similarly, guess the cube roots of \(4913,\; 12167, \;32768.\)

Solution

Video Solution

Reasoning:

By grouping the digits of cube into \(3\) and using Table 7.2

Steps:

(i) \(1331\)

Step 1:

\(1 = \) Group \(2 \)

\(331 =\) Group \(1\)

Step 2: From group \(1\), one’s digit of the cube root can be identified.

\(331=\) One’s digit is \(1\)

 Hence cube root’s one’s digit is \(1\).

Step 3: From group \( 2\), which is \(1\) only.

Hence cube root’s ten’s digit is \(1\).

So we get  \(\sqrt[\mathrm{3}]{1331}=11\)

(ii) \(4913\)

Step 1:

\(4 =\) Group \(2\)

\(913 =\) Group \(1\)

Step 2: From group \(1\), which is \(913\).

\(913\) \(=\) One’s digit is \(3\)

 We know that \(3\) comes at the one’s place of a number only when it’s cube root ends in \(7\). So, we get \(7\) at the one’s place of the cube root. (Refer table 7.2 INFERENCE)

Step 3:From Group 2, which is 4.

\(\begin{align}{{1}^{3}}<4<{{2}^{3}}\end{align} \)

Taking lower limit. Therefore, ten’s digit of cube root is \(1\).

So, we get .\(\begin{align}\sqrt[\mathrm{3}]{4913}=17\end{align}\)

(iii) \(12167\)

Step 1:

\(12 = \) Group \(2\)

\(167 =\) Group \(1\)

Step 2:  From group \(1\), which is \(167\).

 \(167\) \(=\) One’s digit is \(7\)

 We know that \(7\) comes at the one’s place of a number only when it’s cube root ends in \(3\). So, we get \(3\) at the one’s place of the cube root. (Refer table 7.2 INFERENCE)

Step 3: From Group \(2\), which is \(12\)

\({{2}^{3}}<12<{{3}^{3}}\)

Taking the lower limit. Therefore, ten’s digit of cube root is \(2\).

So, we get .\(\sqrt[\mathrm{3}]{12167}=23\)

(iv) \(32768\)

Step 1:

\(32 = \) Group \(2\)

\(768 =\) Group \(1\)

Step 2: From group \(1\), which is \(768\).

\(768\) \(=\) One’s digit is \(8\)

 We know that \(8\) comes at the one’s place of a number only when it’s cube root ends in \(2\). So, we get \(2\) at the one’s place of the cube root. (Refer table 7.2 INFERENCE)

Step 3: From Group \(2\), which is \(32\).

\({{3}^{3}}<32<{{4}^{3}}\)

Taking lower limit. Therefore, ten’s digit of cube root is \( 3\).

 So, we get .\(\sqrt[\mathrm{3}]{32768}=32\)

 

  
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