CBSE Class 8 Maths Chapter 6 Cubes and Cube Roots – Solutions 2025–26

Exercise

1. Which of the following numbers are not perfect cubes?

I. \[{\mathbf{216}}\]

And:The prime factorisation of $216$ is as follows.

\[216\]

\[ = 2 \times 2 \times 2 \times 3 \times 3 \times 3\]

\[ = 23 \times 33\]

Here, as each prime factor is appearing as many times as a perfect multiple of $3$, therefore, $216$ is a perfect cube.

II. \[{\mathbf{128}}\]

Ans: The prime factorisation of \[128\] is as follows

\[128 = 2 \times 2 \times 2 \times {\text{2}} \times 2 \times 2 \times 2\]

Here, each prime factor is not appearing as many times as a perfect multiple of $3$. 

One\[\;2\] is remaining after grouping the triplets of \[\;2\]. 

Therefore, \[128\] is not a perfect cube.

III. \[{\mathbf{1000}}\]

Ans: The prime factorisation of \[1000\] is as follows

\[1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5\]

Here, as each prime factor is appearing as many times as a perfect multiple of $3$,

therefore, \[1000\] is a perfect cube.

IV. \[{\mathbf{100}}\]

Ans: The prime factorisation of \[100\] is as follows. 

\[100 = 2 \times 2 \times 5 \times 5\]

Here, each prime factor is not appearing as many times as a perfect multiple of$3$.

Two \[2s\] and two \[5s\] are remaining after grouping the triplets. 

Therefore, \[100\] is not a perfect cube.

V. \[{\mathbf{46656}}\]

The prime factorisation of \[46656\] is as follows.

\[46656 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3\]

Here, as each prime factor is appearing as many times as a perfect multiple of $3$, therefore, \[46656\] is a perfect cube.

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

I. \[{\mathbf{243}}\]

Ans: \[243 = 3 \times 3 \times 3 \times 3 \times 3\]

Here, two 3s are left which are not in a triplet. To make \[243\] a cube, one more $3$ is required. 

In that case, \[243 \times 3 = 3 \times 3 \times 3 \times 3 \times {\text{3}} \times 3\]

\[ = 729\]

is a perfect cube. 

Hence, the smallest natural number by which \[243\] should be multiplied to make it a perfect cube is $3$.

II. \[{\mathbf{256}}\]

Ans: \[256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2\]

Here, two 2s are left which are not in a triplet. To make \[256\] a cube, one more 2 is required. 

Then, we obtain \[256 \times 2 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2\]

\[ = 512\]

\[512\]is a perfect cube.

III. 72

Ans: \[72 = 2 \times 2 \times 2 \times 3 \times 3\]

Here, two 3s are left which are not in a triplet. To make \[72\] a cube, one more $3$ is required. 

Then, we obtain \[72 \times 3 = 2 \times 2 \times 2 \times 3 \times 3 \times 3\]

\[ = 216\]

\[216\]is a perfect cube. 

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

IV. \[{\mathbf{675}}\]

Ans: \[675 = 3 \times 3 \times 3 \times 5 \times 5\]

Here, two 5s are left which are not in a triplet. To make \[675\] a cube, one more $5$ is required. 

Then, we obtain \[675 \times 5 = 3 \times 3 \times 3 \times 5 \times 5 \times 5\]

\[ = 3375\]

\[3375\] is a perfect cube. 

Hence, the smallest natural number by which \[675\] should be multiplied to make it a perfect cube is $5$.

V. \[{\mathbf{100}}\]

Ans:\[100 = 2 \times 2 \times 5 \times 5\]

Here, two \[2s\] and two \[5s\] are left which are not in a triplet. To make $100$ a cube, we require one more $2$ and one more $5$. 

Then, we obtain \[100 \times 2 \times 5 = 2 \times 2 \times 2 \times 5 \times 5 \times 5\]

\[ = 1000\]

\[1000\]is a perfect cube 

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

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

I. 81

Ans: \[81 = 3 \times 3 \times 3 \times 3\]

Here, one $3$ is left which is not in a triplet. If we divide $81$ by $3$, then it will become a perfect cube. 

Thus, \[\dfrac{{81}}{3} = {\text{2}}7\]

\[ = 3 \times 3 \times 3\]

Whichis a perfect cube. 

Hence, the smallest number by which $81$ should be divided to make it a perfect cube is $3$.

II. \[{\mathbf{128}}\]

Ans: \[128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2\]

Here, one $2$ is left which is not in a triplet. If we divide $128$ by $2$, then it will become a perfect cube. 

Thus, \[\dfrac{{128}}{2} = 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2\]

Which is a perfect cube. 

Hence, the smallest number by which $128$ should be divided to make it a perfect cube is $2$.

III. \[{\mathbf{135}}\]

Ans: \[135 = 3 \times 3 \times 3 \times 5\]

Here, one $5$ is left which is not in a triplet. If we divide \[135\] by $5$, then it will become a perfect cube. 

Thus, \[\dfrac{{135}}{5} = {\text{2}}7 = 3 \times 3 \times 3\]

Which is a perfect cube. 

Hence, the smallest number by which \[135\] should be divided to make it a perfect cube is $5$.

IV. \[{\mathbf{192}}\]

Ans: \[192 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3\]

Here, one $3$ is left which is not in a triplet. If we divide $192$ by $3$, then it will become a perfect cube. 

Thus, \[\dfrac{{192}}{3} = {\text{6}}4 = 2 \times 2 \times 2 \times 2 \times 2 \times 2\]

Which is a perfect cube. 

Hence, the smallest number by which \[192\] should be divided to make it a perfect cube is $3$.

V. \[{\mathbf{704}}\]

Ans: \[704 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 11\]

Here, one \[11\] is left which is not in a triplet. 

If we divide \[704\] by\[11\], then it will become a perfect cube. 

Thus, \[\dfrac{{704}}{{11}} = 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2\]

Which  is a perfect cube.

4. Parikshit makes a cuboid of plasticine of 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?

Ans: Here, some cuboids of size \[5 \times 2 \times 5\]are given.

When these cuboids are arranged to form a cube, the side of this cube so formed will be a common multiple of the sides (i.e., \[5,2,\]and $5$) of the given cuboid. 

LCM of \[5,2,\] and \[5 = 10\]

Let us try to make a cube of $10$ cm. 

For this arrangement, we have to put $2$ cuboids along with its length, $5$ along with its width, and $2$ along with its height.

Total cuboids required according to this arrangement \[ = 2 \times 5 \times 2 = 20\]

With the help of \[20\] cuboids of such measures, a cube is formed as follows.

Conclusion

NCERT Chapter 6 of Class 8 Maths, focusing on Cubes and Cube Roots, is essential for understanding fundamental mathematical concepts. It's important to grasp the properties and patterns of cubes and cube roots, as these are foundational for more advanced topics. Students should focus on understanding how to find cubes and cube roots, recognizing perfect cubes, and applying these concepts in real-life scenarios. Regular practice of class 8 maths 6.1 will ensure a solid grasp of these concepts, aiding in better performance in exams.

Class 8 Maths Chapter 6: Exercises Breakdown

CBSE Class 8 Maths Chapter 6 Other Study Materials

Chapter-Specific NCERT Solutions for Class 8 Maths

Given below are the chapter-wise NCERT Solutions for Class 8 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

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