Find the cubes of:
(i) – 11 (ii) – 12 (iii) – 21
(i) Cube of – 11 is given as: (-11)3 = – 11 x – 11 x – 11 = – 1331
Thus, the cube of 11 is (-1331).
(ii) Cube of – 12 is given as: (-12)3 = – 12 x – 12 x – 12 = – 1728
Thus, the cube of – 12 is (- 1728).
(iii) Cube of – 21 is given as:
(- 21)3 = – 21 x – 21 x – 21 = – 9261
Thus the cube of – 21 is (- 9261).
Which of the following numbers are cubes of negative integers?
(i) – 64
(ii) – 1056
(iii) – 2197
(iv) – 2744
(v) – 42875
In order to check if a negative number is a perfect cube, first check if the corresponding positive integer is a perfect cube. Also, for any positive integer m, – m3 is the cube of – m.
(i) On factorizing 64 into prime factors, we get:
64 = 2 x 2 x 2 x 2 x 2 x 2
On grouping the factors in triples of equal factors, we get:
64 = {2 x 2 x 2} x {2 x 2 x 2}
It is evident that the prime factors of 64 can be grouped into triples of equal factors and no factor is left over. Therefore, 64 is a perfect cube. This implies that – 64 is also a perfect cube.
Now, collect one factor from each triplet and multiply, we get: 2 x 2= 4
This implies that 64 is a cube of 4. Thus, – 64 is the cube of -4.
(ii) On factorising 1056 into prime factors, we get:
1056 = 2 x 2 x 2 x 2 x 2 x 3 x 11
On grouping the factors in triples of equal factors, we get:
1056 = {2 x 2 x 2}x 2 x 2 x 3 x 11
It is evident that the prime factors of 1056 cannot be grouped into triples of equal factors such that no factor is left over.
Therefore, 1056 is not a perfect cube. This implies that – 1056 is not a perfect cube as well.
(iii) On factorising 2197 into prime factors, we get:
2197 =13 x 13 x 13
On grouping the factors in triples of equal factors, we get:
2197 = {13 x 13 x13}
It is evident that the prime factors of 2197 can be grouped into triples of equal factors and no factor is left over. Therefore, 2197 is a perfect cube. This implies that – 2197 is also a perfect cube.
Now, collect one factor from each triplet and multiply, we get 13. This implies that 2197 is a cube of 13. Thus, -2197 is the cube of – 13.
(iv) On factorizing 2744 into prime factors, we get:
2744 = 2 x 2 x 2 x 7 x 7 x 7
On grouping the factors in triples of equal factors, we get: 2744 = {2 x 2 x 2} x {7 x 7 x 7}
It is evident that the prime factors of 2744 can be grouped into triples of equal factors and no factor is left over. Therefore, 2744 is a perfect cube. This implies that – 2744 is also a perfect cube.
Now, collect one factor from each triplet and multiply, we get:
2 x 7 = 14
This implies that 2744 is a cube of 14.
Thus, – 2744 is the cube of – 14.
(v) On factorizing 42875 into prime factors, we get:
42875 = 5 x 5 x 5 x 7 x 7 x 7
On grouping the factors in triples of equal factors, we get:
42875 = {5 x 5 x 5} x {7 x 7 x 7}
It is evident that the prime factors of 42875 can be grouped into triples of equal factors and no factor is left over. Therefore, 42875 is a perfect cube.
This implies that – 42875 is also a perfect cube.
Now, collect one factor from each triplet and multiply, we get: 5 x 7 = 35
This implies that 42875 is a cube of 35. Thus, – 42875 is the cube of – 35.
Show that the following integers are cubes of negative integers. Also find the integer whose cube is the given integer.
(i) – 5832 (ii) – 2744000
In order to check if a negative number is a perfect cube, first check if the corresponding positive integer is a perfect cube. Also, for any positive integer in, – m3 is the cube of – m.
(i) On factorising 5832 into prime factors, we get:
5832 = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
On grouping the factors in triples of equal factors, we get:
5832 = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
It is evident that the prime factors of 5832 can be grouped into triples of equal factors and no factor is left over. Therefore, 5832 is a perfect cube.
This implies that -5832 is also a perfect cube.
Now, collect one factor from each triplet and multiply, we get:
2 x 3 x 3 = 18
This implies that 5832 is a cube of 18.
Thus, – 5832 is the cube of – 18.
(ii) On factorising 2744000 into prime factors, we get:
2744000 = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5 x 5 x 7 x 7 x 7
On grouping the factors in triples of equal factors, we get:
2744000 = {2 x 2 x 2} x {2 x 2 x 2} x {5 x 5 x 5} x {7 x 7 x 7}
It is evident that the prime factors of 2744000 can be grouped into triples of equal factors and no factor is left over. Therefore, 2744000 is a perfect cube. This implies that – 2744000 is also a perfect cube.
Now, collect one factor from each triplet and multiply, we get: 2 x 2 x 5 x 7 = 140
This implies that 2744000 is a cube of 140. Thus, – 2744000 is the cube of – 140.
Find the cube of:
(i) 7/9
(ii) – (8/11)
(iii) 12/7
(iv) –(13/8)
(v) 2(2/5)
(vi) 3(1/4)
(vii) 0.3
(viii) 1.5
(ix) 0.08
(x) 2.1
Find which of the following numbers are cubes of rational numbers?
(i) 27/64
(ii) 125/128
(iii) 0.001331
(iv) 0.04