Answer :

Given that,

Sumanta has made many cubes of length, 1 cm.

This means, volume of each cube = length × length × length

Here, length = 1 cm

⇒ Volume of each cube = 1 × 1 × 1 cm^{3}

⇒ Volume of each cube = 1 cm^{3}

And a single cube out of those many cubes is made by Manami.

Let us assume that Sumanta has made ‘Z’ cubes of length 1 cm, where Z is some natural number.

So, if volume of 1 cube = 1 cm^{3}

⇒ Volume of Z cubes = Z × 1 cm^{3}

⇒ Volume of Z cubes = Z cm^{3}

We have volumes of cube in the options.

**(i).** Let us check for 100.

We need to find cube root of 100. If we are able to find the cube root of 100, then we can say that a big cube is possible in this case.

100 = 2×2× 5× 5

We can see 100 cannot be written as a perfect cube.

∴ 100 ≠ (any integer)^{3}

Since, we have cubes of length 1 cm each.

So, length of the final big cube will also be a natural number.

And cube root of 100 is not a natural number.

**Thus, Manami won’t be able to make a big cube in this case.**

**(ii).** Let us check for 1000.

We need to find the cube root of 1000. If we are able to find the cube root of 1000, then we can say that a big cube is possible in this case.

Cube root of 1000

1000 = 10× 10× 10

⇒ Cube root of 1000 = (1000)^{1/3}

= 10

Since, we have cubes of length 1 cm each.

So, length of the final big cube will also be a natural number.

And cube root of 1000 is a natural number.

**Thus, Manami will be able to make a big cube in this case.**

**(iii).** Let us check for 1331.

We need to find the cube root of 1331. If we are able to find the cube root of 1331, then we can say that a big cube is possible in this case.

Cube root of 1331

1331 = 11× 11× 11

⇒ Cube root of 1331 = (1331)^{1/3}

= 11

Since, we have cubes of length 1 cm each.

So, length of the final big cube will also be a natural number.

And cube root of 1331 is a natural number.

**Thus, Manami will be able to make a big cube in this case.**

**(iv).** Let us check for 1210.

We need to find the cube root of 1210. If we are able to find the cube root of 1210, then we can say that a big cube is possible in this case.

Cube root of 1210

1210 = 2× 5× 11× 11

We can see 1210 cannot be written as a perfect cube.

∴ 1210 ≠ (any integer)^{3}

Since, we have cubes of length 1 cm each.

So, length of the final big cube will also be a natural number.

And cube root of 1210 is not a natural number.

**Thus, Manami will not be able to make a big cube in this case.**

**(v).** Let us check for 3375.

We need to find the cube root of 3375. If we are able to find the cube root of 3375, then we can say that a big cube is possible in this case.

Cube root of 3375

3375 = 15× 15× 15

⇒ Cube root of 3375 = (3375)^{1/3}

= 15

Since, we have cubes of length 1 cm each.

So, length of the final big cube will also be a natural number.

And cube root of 3375 is a natural number.

**Thus, Manami will be able to make a big cube in this case.**

**(vi).** Let us check for 2700.

We need to find the cube root of 2700. If we are able to find the cube root of 2700, we can say that a big cube is possible in this case.

Cube root of 2700

2700 = 3× 3× 3× 2× 2× 5× 5

We can see 2700 cannot be written as a perfect cube.

∴ 2700 ≠ (any integer)^{3}

Since, we have cubes of length 1 cm each.

So, length of the final big cube will also be a natural number.

And cube root of 2700 is not a natural number.

**Thus, Manami will not be able to make a big cube in this case.**

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