Cube Root of 2560
The value of the cube root of 2560 rounded to 6 decimal places is 13.679808. It is the real solution of the equation x^{3} = 2560. The cube root of 2560 is expressed as ∛2560 or 8 ∛5 in the radical form and as (2560)^{⅓} or (2560)^{0.33} in the exponent form. The prime factorization of 2560 is 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5, hence, the cube root of 2560 in its lowest radical form is expressed as 8 ∛5.
 Cube root of 2560: 13.679807573
 Cube root of 2560 in Exponential Form: (2560)^{⅓}
 Cube root of 2560 in Radical Form: ∛2560 or 8 ∛5
1.  What is the Cube Root of 2560? 
2.  How to Calculate the Cube Root of 2560? 
3.  Is the Cube Root of 2560 Irrational? 
4.  FAQs on Cube Root of 2560 
What is the Cube Root of 2560?
The cube root of 2560 is the number which when multiplied by itself three times gives the product as 2560. Since 2560 can be expressed as 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5. Therefore, the cube root of 2560 = ∛(2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5) = 13.6798.
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How to Calculate the Value of the Cube Root of 2560?
Cube Root of 2560 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 = 2560
Let us assume x as 13
[∵ 13^{3} = 2197 and 2197 is the nearest perfect cube that is less than 2560]
⇒ x = 13
Therefore,
∛2560 = 13 (13^{3} + 2 × 2560)/(2 × 13^{3} + 2560)) = 13.68
⇒ ∛2560 ≈ 13.68
Therefore, the cube root of 2560 is 13.68 approximately.
Is the Cube Root of 2560 Irrational?
Yes, because ∛2560 = ∛(2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5) = 8 ∛5 and it cannot be expressed in the form of p/q where q ≠ 0. Therefore, the value of the cube root of 2560 is an irrational number.
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Cube Root of 2560 Solved Examples

Example 1: The volume of a spherical ball is 2560π in^{3}. What is the radius of this ball?
Solution:
Volume of the spherical ball = 2560π in^{3}
= 4/3 × π × R^{3}
⇒ R^{3} = 3/4 × 2560
⇒ R = ∛(3/4 × 2560) = ∛(3/4) × ∛2560 = 0.90856 × 13.67981 (∵ ∛(3/4) = 0.90856 and ∛2560 = 13.67981)
⇒ R = 12.42893 in^{3} 
Example 2: What is the value of ∛2560 + ∛(2560)?
Solution:
The cube root of 2560 is equal to the negative of the cube root of 2560.
i.e. ∛2560 = ∛2560
Therefore, ∛2560 + ∛(2560) = ∛2560  ∛2560 = 0

Example 3: Given the volume of a cube is 2560 in^{3}. Find the length of the side of the cube.
Solution:
Volume of the Cube = 2560 in^{3} = a^{3}
⇒ a^{3} = 2560
Cube rooting on both sides,
⇒ a = ∛2560 in
Since the cube root of 2560 is 13.68, therefore, the length of the side of the cube is 13.68 in.
FAQs on Cube Root of 2560
What is the Value of the Cube Root of 2560?
We can express 2560 as 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5 i.e. ∛2560 = ∛(2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5) = 13.67981. Therefore, the value of the cube root of 2560 is 13.67981.
Why is the Value of the Cube Root of 2560 Irrational?
The value of the cube root of 2560 cannot be expressed in the form of p/q where q ≠ 0. Therefore, the number ∛2560 is irrational.
If the Cube Root of 2560 is 13.68, Find the Value of ∛2.56.
Let us represent ∛2.56 in p/q form i.e. ∛(2560/1000) = 13.68/10 = 1.37. Hence, the value of ∛2.56 = 1.37.
How to Simplify the Cube Root of 2560/343?
We know that the cube root of 2560 is 13.67981 and the cube root of 343 is 7. Therefore, ∛(2560/343) = (∛2560)/(∛343) = 13.68/7 = 1.9543.
What is the Cube of the Cube Root of 2560?
The cube of the cube root of 2560 is the number 2560 itself i.e. (∛2560)^{3} = (2560^{1/3})^{3} = 2560.
What is the Cube Root of 2560?
The cube root of 2560 is equal to the negative of the cube root of 2560. Therefore, ∛2560 = (∛2560) = (13.68) = 13.68.
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