Cubes and Cube Roots

Mathematics

Mathematics

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Is 1372 a perfect cube? If not, find the smallest natural number by which 1372 must be multipled so that the product is a perfect cube.

We have 1372 = 2 x 2 x 7 x 7 x 7

Since, the prime factor 2 does not appear in a group of triples.

∴ 1372 is not a perfect cube.

Obviously, to make it a perfect cube we need one more 2 as its factor.

i.e. [1372] x 2 = [2 x 2 x 7 x 7 x 7] x 2

or 2744 = 2 x 2 x 2 x 7 x 7 x 7

which is a perfect cube.

Thus, the required smallest number = 2.

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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 (v) 704

(i) We have 81 = 3 x 3 x 3 x 3

Grouping the prime factors of 81 into triples, we are left with 3.

∴ 81 is not a perfect cube

Now, [81] 3= [3 x 3 x 3 x 3] 3

or 27 = 3 x 3 x 3

i.e. 27 is a perfect cube

Thus, the required smallest number is 3

(ii) we have 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2

Grouping the prime factors of 128 into triples, we are left with 2

∴ 128 is not a perfect cube

Now, [128] 2 = [2 x 2 x 2 x 2 x 2 x 2 x 2]2

or 64 = 2 x 2 x 2 x 2 x 2 x 2

i.e. 64 is a perfect cube

∴ the smallest required number is 2.

(iii) we have 135 = 3 x 3 x 3 x 5

Grouping the prime factors of 135 into triples, we are left over with 5.

∴ 135 is not a perfect cube

Now, [135]5 = [ 3 x 3 x 3 x 5] 5

or 27 = 3 x 3 x 3

i.e. 27 is a perfect cube.

Thus, the required smallest number is 5

(iv) We have 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3

Grouping the prime factors of 192 into triples, 3 is left over.

∴ 192 is not a perfect cube.

Now, [192] 3= [2 x 2 x 2 x 2 x 2 x 2 x 3]3

or 64 = 2 x 2 x 2 x 2 x 2 x 2

i.e. 64 is a perfect cube.

Thus, the required smallest number is 3.

(v) We have 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11

Grouping the prime factors of 704 into triples, 11 is left over

∴ [704]11 = [2 x 2 x 2 x 2 x 2 x 2 x 11]11

or 64 = 2 x 2 x 2 x 2 x 2 x 2

i.e. 64 is a perfect cube

Thus, the required smallest number is 11.

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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

(i) We have 243 = 3 x 3 x 3 x 3 x 3

The prime factor 3 is not a group of three.

∴ 243 is not a perfect cube.

Now, [243] x 3 = [3 x 3 x 3 x 3 x 3] x 3

or, 729, = 3 x 3 x 3 x 3 x 3 x 3** **Now, 729 becomes a [perfect cube

Thus, the smallest required number to multipkly 243 to make it a perfect cube is 3.

(ii) We have 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

Grouping the prime factors of 256 in triples, we are left over with 2 x 2.

∴ 256 is not a perfect cube.

Now, [256] x 2 = [2 x 2 x 2 x 2 x 2 x 2 x 2 x 2] x 2

or, 512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

i.e. 512 is a perfect cube.

thus, the required smallest number is 2.

(iii) we have 72 = 2 x 2 x 2 x 3 x 3

Grouping the prime factors of 72 in triples, we are left over with 3 x 3

∴ 72 is not a perfect cube.

Now, [72] x 3 = [2 x 2 x 2 x 3 x 3] x 3

or, 216 = 2 x 2 x 2 x 3 x 3 x 3

i.e. 216 is a perfect cube

∴ The smallest number required to multiply 72 to make it a perfect cube is 3.

(iv) We have 675 = 3 x 3 x 3 x 5 x 5

Grouping the prime factors of 675 to triples, we are left over with 5 x 5

∴ 675 is not a perfect cube.

Now, [675] x 5 = [3 x 3 x 3 x 5 x 5] x 5

Now, 3375 is a perfect cube

Thus, the smallest required number to multiply 675 such that the new number is a perfect cube is 5.

(v) We have 100 = 2 x 2 x 5 x 5

The prime factor are not in the groups of triples.

∴ 100 is not a perfect cube.

Now, [100] x 2 x 5 = [2 x 2 x 5 x 5] x 2 x 5

or, [100] x 10 = 2 x 2 x 2 x 5 x 5 x 5

1000 = 2 x 2 x 2 x 5 x 5 x 5

Now, 1000 is a perfect cube

Thus, the required smallest number is 10

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Which of the following are perfect cubes?

1. 400 2. 3375 3. 8000 4. 15625

5. 9000 6. 6859 7. 2025 8. 10648

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You are told that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.

(i) Separating the given number (1331) into two groups :

1331 1 and 331

∵ 331 end in 1

∴ Unit's digit of the cube root = 1

∵ 1^{3 }= 1 and

∴ Ten's digit of the cube root = 1

∴

(ii) Separating the given number (4913) in two groups:

4913 4 and 913

Unit's digit:

∵ Unit's digit in 913 is 3

∴ Unit's digit of the cube root = 7

[7^{3 }= 343 : which ends in 3]

Ten's digit:

∵ 1^{3 }= 1, 2^{3 } = 8

and 1 < 4 < 8

i.e. 1^{3 } < 4 < 2^{3 }

∴ Then ten's digit of the cube root is 1.

∴

(iii) Separatibng 12167 in two groups:

1216712 and 167

Unit's digit :

∵ 167 is ending in 7 and cube of a number ending in 3 ends in 7

∴ The unit's digit of the cube root = 3

Ten's digit

∵ 2^{3 } = 8 adn 3^{3 }= 27

Also, 8 < 12 < 27

or, 2^{3 }< 12 < 3^{2 }∴ The tens digit of the cube root can be 2.

Thus,

(iv) separating 32768 in two groups:

32768 32 and 786

Unit's digit:

768 will guess the unit's digit in the cube root.

∵ 768 ends in 8.

∴ Unit's digit in the cube root = 2

Ten's digit:

∵ 3^{3 }= 27 and 4^{3 }- 64

Also, 27 < 32 < 64

or, 3^{3 }< 32 < 4^{3 }

∴ The ten's digit of the cube root = 3

Thus,

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Check which of the following are perfect cubes.

(i) 2700 (ii) 16000 (iii) 64000 (iv) 900

(v) 125000 (vi) 36000 (vii) 21600 (viii) 10000

(ix) 27000000 (x) 1000

What pattern do you observe in these perfect cubes?

(i) We have 2700 = 2 x 2 x 3 x 3 x 3 x 5 x 5

We do not get complete triples of prime factors, i.e. 2 x 2 and 5 x 5 are left over.

(ii) we have 1600 = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5

After grouping id 3's we get 5 x 5 which is ungrouped in triples.

(iii) We have 64000 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5 x 5

Since, we get groups of triples.

(iv) We have 900 = 2 x 2 x 3 x 3 x 5 x 5

which are ungrouped in triples.

(v) We have 125000 = 2 x 2 x 2 x 5 x 5 x 5 x 5 x 5 x 5

As we get all the prime factors in the group of triples.

(vi) we have 36000 = 2 x 2 x 2 x 2 x 2 x 3 x 3 x 5 x 5 x 5

While grouping the prime factors of 36000 in triples, we are left over with 2 x 2 and 3 x 3

(vii) We have 21600 = 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 5 x 5

While grouping the prime factors of 21600 in triples, we are left with 2 x 2 and 5 x 5.

(viii) we have 10000 = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5

While grouping the prime factors into triples, we are left over with 2 and 5.

(ix) we have 27000000 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 5 x 5 x 5 x 5 x 5 x 5

Since, all the prime factors of 27000000 appear in groups of triples.

(x) We have 1000 = 2 x 2 x 2 x 5 x 5 x 5

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