What is a cube root of 1372?

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Here we will show you how to get the factors of cube root of 1372 (factors of ∛1372). We define factors of cube root of 1372 as any integer (whole number) or cube root that you can evenly divide into cube root of 1372. Furthermore, if you divide ∛1372 by a factor of ∛1372, it will result in another factor of ∛1372.

First, we will find all the cube roots that we can evenly divide into cube root of 1372. We do this by finding all the factors of 1372 and add a radical (∛) to them like this:

∛1, ∛2, ∛4, ∛7, ∛14, ∛28, ∛49, ∛98, ∛196, ∛343, ∛686, and ∛1372

Next, we will find all the integers that we can evenly divide into cube root of 1372. We do that by first identifying the perfect cube roots from the list above:

∛1 and ∛343

Then, we take the cube root of the perfect cube roots to get the integers that we can evenly divide into cube root of 1372.

1 and 7

Factors of cube root of 1372 are the two lists above combined. Thus, factors of cube root of 1372 (cube roots and integers) are as follows:

1, 7, ∛1, ∛2, ∛4, ∛7, ∛14, ∛28, ∛49, ∛98, ∛196, ∛343, ∛686, and ∛1372

Like we said above, cube root of 1372 divided by any of its factors, will result in another of its factors. Therefore, if you divide ∛1372 by any of factors above, you will see that it results in one of the other factors.

What can you do with this information? For one, you can get cube root of 1372 in its simplest form. Cube root of 1372 simplified is the largest integer factor times the cube root of 1372 divided by the largest perfect cube root. Thus, here is the math to get cube root of 1372 in its simplest radical form:

∛1372
= 7 × (∛1372 ÷ ∛343)
= 7∛4

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Factors of Cube Root of 1373
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Here we will show you the answer to the cube root of 1372 in decimal form and in its simplest radical form. But before that, we start by showing you the cube root of 1372 written in mathematical terms:

∛1372

The √ is called the radical symbol, the little 3 is the index which means cube, and 1372 is the radicand.

The cube root of 1372 is a number (n) that when multiplied by itself twice will equal 1372. In other words, the cube root of 1372 is the number (n) in the following equation:

n × n × n = 1372

∛1372 is the same as 1372⅓. Therefore, to solve the problem in Excel or Numbers, you can enter 1372^(1/3) to get the answer to the cube root of 1372. However, we used a scientific calculator and typed in 1372 and then pressed the [∛x] button. We got the following answer:

∛1372 ≈ 11.1118

You can simplify the cube root of 1372 if you can make the radicand smaller. This is called the cube root of 1372 in its simplest radical form. The answer to the cube root 1372 in its simplest form is displayed below:

∛1372 = 7∛4

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Cube root of 1373
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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.

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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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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.

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(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

∵ 13 = 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

[73 = 343 : which ends in 3]

Ten's digit:
∵ 13 = 1, 23  = 8
and 1 < 4 < 8
i.e. 13  <  4 < 23 
∴ Then ten's digit of the cube root is 1.

∴  

(iii) Separatibng 12167 in two groups:
     12167

12 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
∵ 23  = 8 adn 33 = 27
Also,  8 < 12 < 27
or,  23 < 12 < 32  ∴ 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:
∵ 33 = 27 and 43 - 64
Also, 27 < 32 < 64
or, 33 < 32 < 43 

∴ 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?

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(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.

∴ 2700 is not a perfect cube.
(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.
∴ 1600 is not a perfect cube.
(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.

∴ 64000 is a perfect cube.
(iv) We have 900 = 2 x 2 x 3 x 3 x 5 x 5
which are ungrouped in triples.

∴ 900 is not a perfect cube.
(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.
∴ 125000 is a perfect cube
(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
∴ 36000 is not a perfect cube
(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.
∴ 21600 is not a perfect cube
(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.

∴ 10000 is not a perfect cube
(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.
∴ 27000000 is a perfect cube.
(x) We have 1000 = 2 x 2 x 2 x 5 x 5 x 5

∴ 1000 is a perfect cube.

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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

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(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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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

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(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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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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What is the cube of 1372?

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