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
Gii) 72
(v) 675
(v) 100
​

Answer 1

Answer:

Thus, the smallest required number tomultiply 243 to make it a perfect cube is 3. Grouping the prime factors of 256 in triples, we are left over with 2 × 2. ∴ 256is not a perfect cube. i.e. 512 is a perfect cube.

Step-by-step explanation:

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Answer 2

Answer:

1 Prime factors of 243 =

Here 3 do not appear in 3’s group.

Therefore, 243 must be multiplied by 3 to make it a perfect cube.

2. Prime factors of 256 = 2*2*2*2*2*2*2*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.

3. Prime factors of 72 = 2×2×2×3×3

Here 3 does not appear in 3’s group.

Therefore, 72 must be multiplied by 3 to make it a perfect cube.

4. Prime factors of 675 = 3×3×3×5×5

Here factor 5 does not appear in 3’s group.

Therefore 675 must be multiplied by 3 to make it a perfect cube.

5. Prime factors of 100 = 2×2×5×5

Here factor 2 and 5 both do not appear in 3’s group.

Therefore 100 must be multiplied by 2×5= 10 to make it a perfect cube.