# Exercise 7.1 Cubes and Cube Roots- NCERT Solutions Class 8

Rational Numbers

Ex.7.1

## Chapter 7 Ex.7.1 Question 1

Which of the following numbers are not perfect cubes?

(i) \(216\)

(ii) \(128\)

(iii) \(1000\)

(iv) \(100\)

(v) \(46656\)

**Solution**

**Video Solution**

**What is unknown?**

To find the numbers which are not perfect cubes.

**Reasoning: **

A number is a prefect cube only when each factor in the prime factorization is grouped in triples.

**Steps: **

(i)

\[\begin{align}216 &= {2 \times 2 \times 2} \times {3 \times 3 \times 3} \\&= {2^3} \times {3^3}\\& ={{(2\times 3)}^{3}} \\& ={{6}^{3}} \\\end{align}\]

\(\therefore 216\) is a perfect cube

(ii)

\[\begin{align}128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \end{align}\]

\(128 = {2^3} \times {2^3} \times 2\) one of the \(2\) is not grouped in triples

\(\therefore \;128\) is not a perfect cube

(iii)

\[\begin{align}1000 &= 2 \times 2 \times 2 \times 5 \times 5 \times 5\\

1000 &= {2^3} \times {5^3}\end{align}\]

\(\therefore\;1000\) is a perfect cube

(iv)

\[\begin{align}100&= 2 \times 2 \times 5 \times 5\\&= {2^2} \times {5^2}\end{align}\]

Both \(2\) and \(5\) are not grouped in triples

\(\therefore \;100\) is not a perfect cube.

(iv)

\(\begin{align}46656 &= \left(\begin{array} \ \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{array} \right) \\ &= {2^3} \times {2^3} \times {3^3} \times {3^3}

\end{align}\)

\(\therefore\;46656\) is a perfect cube

## Chapter 7 Ex.7.1 Question 2

Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.

(i) \(243\)

(ii) \(256\)

(iii) \(72\)

(iv) \(675\)

(v) \(100\)

**Solution**

**Video Solution**

**What is unknown?**

To find the smallest number by which the given number must be multiplied to obtain a perfect cube.

**Reasoning: **

A number is a perfect cube only when each factor in the prime factorization is grouped in triples.Using this concept, the smallest number can be identified.

**Steps:**

(i)

\[\begin{align}243 &= \underline {3 \times 3 \times 3} \times 3 \times 3\\243 &= {3^3} \times 3\end{align}\]

Here, one of the \(3'\rm{s}\) is not a triplet.To make it as a triplet, we need to multiply by \(3\)

In that case,

\(\begin{align}243 \times 3 &= \underline {3 \times 3 \times 3} \times \underline {3 \times 3 \times 3} \\&= {3^3} \times {3^3} \\&= {9^3} = 729\end{align}\)

is a perfect cube.

Hence, the smallest natural number by which \(243\) should be multiplied to make a perfect cube is \(3\).

(ii)

\[\begin{align}256 &= \underline {2 \times 2 \times 2} \times \underline {2 \times 2 \times 2}\\& \times 2 \times 2\\256 &= {2^3} \times {2^3} \times {2^2}\end{align}\]

Here one of the \( 2’\rm{s}\) is not a triplet. To make it as a triplet, we need to multiply by \(2.\)

In that case,

\(\begin{align}256 \times 2 &= \underline {2 \times 2 \times 2} \times \underline {2 \times 2 \times 2} \\&\times \underline {2 \times 2 \times 2} \\&= {2^3} \times {2^3} \times {2^3} \\&= {8^3} = 512\end{align}\)

is a perfect cube.

Hence, the smallest natural number by which \(256\) should be multiplied to make a perfect cube is \(2\).

(iii)

\[\begin{align}72 &= \underline {2 \times 2 \times 2} \times 3 \times 3\\72 &= {2^3} \times {3^2}\end{align}\]

Here,on of the \(3\)'s is not a triplet. To make it as a triplet, we need to multiply by \(3.\)

In that case,

\(\begin{align}72 \times 3 &= \underline {2 \times 2 \times 2} \times \underline {3 \times 3 \times 3} \\&= {2^3} \times {3^3} \\&= {6^3} = 216\end{align}\)

is a perfect cube.

Hence, the smallest natural number by which 72 should be multiplied to make a perfect cube is 3.

(iv)

\[\begin{align}675 &= 5 \times 5 \times \underline {3 \times 3 \times 3} \\675 &= {5^2} \times {3^3}\end{align}\]

Here, one of the \(5\)'s is not a triplet. To make it as a triplet, we need to multiply by \(5.\)

In that case,

\(\begin{align}675 &= \underline {5 \times 5 \times 5} \times \underline {3 \times 3 \times 3} \\&= {5^3} \times {3^3} \\&= {15^3} = 3375\end{align}\)

is a perfect cube.

Hence, the smallest natural number by which \(675\) should be multiplied to make a perfect cube is \(5\).

(v)

\[\begin{align}100 &= 2 \times 2 \times 5 \times 5\\100 &= {2^2} \times {5^2}\end{align}\]

Here both the prime factors are not triplets. To make them triplets we need to multiply by one more \(2\) and \(5.\)

In that case,

\(\begin{align}100 &=\, \underline {2 \times 2 \times 2} \times \underline {5 \times 5 \times 5} \\&= {2^3} \times {5^3} \\&= {10^3} = 1000\end{align}\)

is a perfect cube.

Hence, the smallest natural number by which 100 should be multiplied to make a perfect cube is \(2 \times 5 = 10.\)

## Chapter 7 Ex.7.1 Question 3

Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.

(i) \(81\)

(ii) \(128\)

(iii) \(135\)

(iv) \(192\)

(v) \(704\)

**Solution**

**Video Solution**

**What is unknown?**

To find the smallest number by which a given number must be divided to obtain a perfect cube.

**Reasoning: **

A number is a perfect cube only when each factor in the prime factorization is grouped in triples. Using this concept smallest number to be multiplied can be obtained.

(i)

\[\begin{align}81 &= \underline {3 \times 3 \times 3} \times 3\\81 &= {3^3} \times 3\end{align}\]

Here, the prime factor \(3\) is not present as triples.

Hence, we divide by \(81\) by \(3,\) so that the obtained number becomes a perfect cube.

Thus,

\(\begin{align}81 \div 3 = 27 = {3^3}\end{align}\) is a perfect cube.

Hence the smallest number by which \(81\) should be divided to make a perfect cube is \(3\).

(ii)

\[\begin{align}128& = \underline {2 \times 2 \times 2} \times \underline {2 \times 2 \times 2} \times 2\\128 &= {2^3} \times {2^3} \times 2\end{align}\]

Here,the prime factors \(2\) is not present as triples.

Hence, we divide \( 128\) by \(2,\) so that the obtained number becomes a perfect cube.

\(\begin{align}128 \div 2 = 64 = {2^3} \times {2^3} = {4^3}\end{align}\) is a perfect cube.

Hence the smallest number by which \(128\) should be divided to make a perfect cube is \(2\).

(iii)

\[\begin{align}135 &= 5 \times \underline {3 \times 3 \times 3} \\135 &= {5^1} \times {3^3}\end{align}\]

Here, the prime factors \(5\) is not present as triples.

Hence, we divide \(135\) by \(5,\) so that the obtained number becomes a perfect cube.

\(\begin{align}135 \div 5 = 27 = {3^3}\end{align}\) is a perfect cube.

Hence the smallest number by which \(135\) should be divided to make a perfect cube is \(5\).

(iv)

\[\begin{align}192 &= \underline {2 \times 2 \times 2} \times \underline {2 \times 2 \times 2} \times 3\\192 &= {2^3} \times {2^3} \times 3\end{align}\]

Here,the prime factors \(3\) is not present as triples.

Hence, we divide \(192\) by \(3,\) so that the obtained number becomes a perfect cube.

\(\begin{align}192 \div 3 = 64 = {2^3} \times {2^3} = {4^3}\end{align}\) is a perfect cube.

Hence the smallest number by which \(192\) should be divided to make a perfect cube is \(3\).

(v)

\[\begin{align}704 &= \underline {2 \times 2 \times 2} \times \underline {2 \times 2 \times 2} \times 11\\704 &= {2^3} \times {2^3} \times 11\end{align}\]

Here, the prime factors \(11\) is not present as triples.

Hence, we divide \(704\) by \(11,\) so that the obtained number becomes a perfect cube.

\(\begin{align}704 \div 11 = 64 = {2^3} \times {2^3} = {4^3}\end{align}\) is a perfect cube.

Hence the smallest number by which \(704\) should be divided to make a perfect cube is \(11\).

## Chapter 7 Ex.7.1 Question 4

Parikshit makes a cuboid of plasticine of sides \(5 \,\rm{cm},\) \(2 \,\rm{cm,}\) \(5\, \rm{cm.}\) How many such cuboids will he need to form a cube?

**Solution**

**Video Solution**

**What is known?**

Dimensions of cuboid \(5\;{\rm{cm}}\, \times\,2\;{\rm{cm}}\,\times \,5\;\rm{cm}.\)

**What is unknown?**

To find out the number of cuboids to form a cube.

**Reasoning: **

Number of cuboids required

\[\begin{align}=\frac{{{\text{Volume of cube}}}}{{{\text{Volume of cuboid}}}}\end{align}\]

**Steps:**

Volume of cuboid

\(\begin{align}&={\rm{length}}\! \times\! {\rm{breadth}} \\&\times {\rm{height}}\\

&= 5 \times 2 \times 5\\&= {5^2} \times {2^1}{\rm{c}}{{\rm{m}}^3}\end{align}\)

To make the volume of cuboid as a cube number we need to multiply it by \(5 \times 2 \times 2\)

\(\begin{align}&\text {Newly formed cube}\\&= {5^2} \times {2^1} \times 5\; \times 2 \times 2\\&= {5^3} \times {2^3}{\rm{c}}{{\rm{m}}^3}\end{align}\)

\(\begin{align}&\text {Number of cuboids required} \\&=\frac{5^{3} \times 2^{3}}{5^{2} \times 2}\\&=\frac{5 \times \not5 \times \not5 \times \not2 \times 2 \times 2}{\not 5 \times \not 5 \times \not2}\end{align}\)

\(\therefore \;\) Number of cuboids required to make a cube \(=\) \(\begin{align}5\times 2\times 2=20\end{align}\)

\(20\) cuboids required.