Cube Root of 1125
The value of the cube root of 1125 rounded to 5 decimal places is 10.40042. It is the real solution of the equation x^{3} = 1125. The cube root of 1125 is expressed as ∛1125 or 5 ∛9 in the radical form and as (1125)^{⅓} or (1125)^{0.33} in the exponent form. The prime factorization of 1125 is 3 × 3 × 5 × 5 × 5, hence, the cube root of 1125 in its lowest radical form is expressed as 5 ∛9.
 Cube root of 1125: 10.400419115
 Cube root of 1125 in Exponential Form: (1125)^{⅓}
 Cube root of 1125 in Radical Form: ∛1125 or 5 ∛9
1.  What is the Cube Root of 1125? 
2.  How to Calculate the Cube Root of 1125? 
3.  Is the Cube Root of 1125 Irrational? 
4.  FAQs on Cube Root of 1125 
What is the Cube Root of 1125?
The cube root of 1125 is the number which when multiplied by itself three times gives the product as 1125. Since 1125 can be expressed as 3 × 3 × 5 × 5 × 5. Therefore, the cube root of 1125 = ∛(3 × 3 × 5 × 5 × 5) = 10.4004.
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How to Calculate the Value of the Cube Root of 1125?
Cube Root of 1125 by Halley's Method
Its formula is ∛a ≈ x ((x^{3} + 2a)/(2x^{3} + a))
where,
a = number whose cube root is being calculated
x = integer guess of its cube root.
Here a = 1125
Let us assume x as 10
[∵ 10^{3} = 1000 and 1000 is the nearest perfect cube that is less than 1125]
⇒ x = 10
Therefore,
∛1125 = 10 (10^{3} + 2 × 1125)/(2 × 10^{3} + 1125)) = 10.4
⇒ ∛1125 ≈ 10.4
Therefore, the cube root of 1125 is 10.4 approximately.
Is the Cube Root of 1125 Irrational?
Yes, because ∛1125 = ∛(3 × 3 × 5 × 5 × 5) = 5 ∛9 and it cannot be expressed in the form of p/q where q ≠ 0. Therefore, the value of the cube root of 1125 is an irrational number.
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Cube Root of 1125 Solved Examples

Example 1: The volume of a spherical ball is 1125π in^{3}. What is the radius of this ball?
Solution:
Volume of the spherical ball = 1125π in^{3}
= 4/3 × π × R^{3}
⇒ R^{3} = 3/4 × 1125
⇒ R = ∛(3/4 × 1125) = ∛(3/4) × ∛1125 = 0.90856 × 10.40042 (∵ ∛(3/4) = 0.90856 and ∛1125 = 10.40042)
⇒ R = 9.44941 in^{3} 
Example 2: Find the real root of the equation x^{3} − 1125 = 0.
Solution:
x^{3} − 1125 = 0 i.e. x^{3} = 1125
Solving for x gives us,
x = ∛1125, x = ∛1125 × (1 + √3i))/2 and x = ∛1125 × (1  √3i))/2
where i is called the imaginary unit and is equal to √1.
Ignoring imaginary roots,
x = ∛1125
Therefore, the real root of the equation x^{3} − 1125 = 0 is for x = ∛1125 = 10.4004. 
Example 3: Given the volume of a cube is 1125 in^{3}. Find the length of the side of the cube.
Solution:
Volume of the Cube = 1125 in^{3} = a^{3}
⇒ a^{3} = 1125
Cube rooting on both sides,
⇒ a = ∛1125 in
Since the cube root of 1125 is 10.4, therefore, the length of the side of the cube is 10.4 in.
FAQs on Cube Root of 1125
What is the Value of the Cube Root of 1125?
We can express 1125 as 3 × 3 × 5 × 5 × 5 i.e. ∛1125 = ∛(3 × 3 × 5 × 5 × 5) = 10.40042. Therefore, the value of the cube root of 1125 is 10.40042.
What is the Cube Root of 1125?
The cube root of 1125 is equal to the negative of the cube root of 1125. Therefore, ∛1125 = (∛1125) = (10.4) = 10.4.
If the Cube Root of 1125 is 10.4, Find the Value of ∛1.125.
Let us represent ∛1.125 in p/q form i.e. ∛(1125/1000) = 10.4/10 = 1.04. Hence, the value of ∛1.125 = 1.04.
Is 1125 a Perfect Cube?
The number 1125 on prime factorization gives 3 × 3 × 5 × 5 × 5. Here, the prime factor 3 is not in the power of 3. Therefore the cube root of 1125 is irrational, hence 1125 is not a perfect cube.
How to Simplify the Cube Root of 1125/343?
We know that the cube root of 1125 is 10.40042 and the cube root of 343 is 7. Therefore, ∛(1125/343) = (∛1125)/(∛343) = 10.4/7 = 1.4857.
What is the Cube of the Cube Root of 1125?
The cube of the cube root of 1125 is the number 1125 itself i.e. (∛1125)^{3} = (1125^{1/3})^{3} = 1125.
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