Question

Exercise 7.1
1. Which of the following numbers are not perfect
cubes?
(1) 216 (ii) 128 (iii) 1000 (iv) 100
(v) 46656
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
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​

Answer 1

Step-by-step explanation:

1)Now, 3375 is a perfect cube. Thus, the smallest required number to multiply 675 such that the new number perfect cube is 5. The prime factor are not in the groups of triples. ∴100 is not a perfect cube.

Answer 2

. (i) 216
(ii) 128
Prime factors of 216 = 2 x 2 x 2 x 3 x 3 x 3
Here all factors are in groups of 3’s (in triplets) Therefore, 216 is a perfect cube number.
Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2 Here one factor 2 does not appear in a 3’s group. Therefore, 128 is not a perfect cube.
Prime factors of 1000 = 2 x 2 x 2 x 3 x 3 x 3 Here all factors appear in 3’s group. Therefore, 1000 is a perfect cube.
2 216 2 108 2 54 3 27 39 33
1
2 128 2 64 2 32 2 16 28 24 22
1
2 1000 2 500 2 250 5 125 5 25 55

(iii) 1000
(iv) 100
2 100 Prime factors of 100 = 2 x 2 x 5 x 5 250
Here all factors do not appear in 3’s group. 525
Therefore, 100 is not a perfect cube.
55 1

(v) 46656
Prime factors of 46656 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3 Here all factors appear in 3’s group.
Therefore, 46656 is a perfect cube.
2 46656
2 23328
2 11664
2 5832
2 2916
2 1458
3 729
3 243
3 81
3 27
39
33
1
3 243 381 327 39 33
1
2 256 2 128 2 64 2 32 2 16 28 24 22
1
272 236 218 39 33
1
2. (i) 243
Prime factors of 243 = 3 x 3 x 3 x 3 x 3
Here 3 does not appear in 3’s group.
Therefore, 243 must be multiplied by 3 to make it a perfect cube.
(ii) 256
Prime factors of 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
Here one factor 2 is required to make a 3’s group.
Therefore, 256 must be multiplied by 2 to make it a perfect cube.
(iii) 72
Prime factors of 72 = 2 x 2 x 2 x 3 x 3
Here 3 does not appear in 3’s group.
Therefore, 72 must be multiplied by 3 to make it a perfect cube.

(iv) 675
Prime factors of 675 = 3 x 3 x 3 x 5 x 5
Here factor 5 does not appear in 3’s group.
Therefore 675 must be multiplied by 3 to make it a perfect cube.
(v) 100
Prime factors of 100 = 2 x 2 x 5 x 5
Here factor 2 and 5 both do not appear in 3’s group. Therefore 100 must be multiplied by 2 x 5 = 10 to make it a perfect cube.
3. (i) 81
Prime factors of 81 = 3 x 3 x 3 x 3
Here one factor 3 is not grouped in triplets.
Therefore 81 must be divided by 3 to make it a perfect cube.
(ii) 128
Prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
Here one factor 2 does not appear in a 3’s group.
Therefore, 128 must be divided by 2 to make it a perfect cube.
(iii) 135
Prime factors of 135 = 3 x 3 x 3 x 5
Here one factor 5 does not appear in a triplet.
Therefore, 135 must be divided by 5 to make it a perfect cube.
3 675 3 225 3 75 5 25 55
1
2 100 250 525 55
1
381 327 39 33
1
2 128 2 64 2 32 2 16 28 24 22
1
3 135 345 315 55
1

(iv) 192
Prime factors of 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3
Here one factor 3 does not appear in a triplet.
Therefore, 192 must be divided by 3 to make it a perfect cube.
(v) 704
Prime factors of 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11
Here one factor 11 does not appear in a triplet.
Therefore, 704 must be divided by 11 to make it a perfect cube.