NCERT Class 8 Mathematics Solutions: Chapter 7 – Cubes and Cube Roots Exercise 7.1 Part 2
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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
Answer:
(i) 243
3 | 243 |
3 | 81 |
3 | 27 |
3 | 9 |
3 | 3 |
1 | |
Prime factors of
To make 243 a cube, one more 3 is required.
Hence, the number 243 multiply by 3 to obtain a perfect cube.
(ii) 256
2 | 256 |
2 | 128 |
2 | 64 |
2 | 32 |
2 | 16 |
2 | 8 |
2 | 4 |
2 | 2 |
1 |
Prime factors of 256 =
To make 256 a cube, one more 2 is required.
Hence, the number 256 multiply by 2 to obtain a perfect cube.
(iii) 72
2 | 72 |
2 | 36 |
2 | 18 |
3 | 9 |
3 | 3 |
1 |
Prime factors of
To make 72 a cube, one more 3 is required.
Hence, the number 72 multiply by 3 to obtain a perfect cube.
(iv) 675
3 | 675 |
3 | 225 |
3 | 75 |
5 | 25 |
5 | 5 |
1 |
Prime factors of
To make 675 a cube, one more 5 is required.
Hence, the number 675 multiply by 5 to obtain a perfect cube.
(v) 100
2 | 100 |
2 | 50 |
5 | 25 |
5 | 5 |
1 |
Prime factors of
To make 100 a cube, we require one more 2 and one more 5.
Hence, the number 100 multiply by to obtain a perfect cube.