10+ Python Cube Root Finder Program Examples

This post contains a total of 10+ Hand-Picked Python Cube Root Finder Program Examples with Source Code. All these Cube root programs are made using Python Programming Language.

You can use the source code of these examples with credits to the original owner.

Python Cube Root Finder Programs

1. By David Akhihiero

Made by David Akhihiero. Simple program to find the cube root of a number. Source

```1.4142156862745097
2.1544177077840487
2.1544177077840487```
``````precision = 0.0001

def ab(x):
if x < 0:
return -x
else:
return x

def sqrt(x, y = 1):
if ab(x / y - y) <= precision:
return y
else:
return sqrt(x, (x / y + y)/2)

print(sqrt(2))

def cbrt(x, y = 1):
if x < 0:
return -1 * cbrt(ab(x))
if ab((x / (y * y)) - y) <= precision:
return y
else:
return cbrt(x, (y + sqrt(x / y)) / 2)

print(cbrt(10))
print(cbrt(10))``````

2. By Sweet Devil

Made by Sweet Devil. Python program to find the cube root of a number N to a precision of 0.0001, with library functions. Source

```5
Cube root from 5 is equal to 1.7100```
``````n = int(input())
x,e = 1,(1-n)/3
while e>1e-5 or e<-1e-5: x,e = x-e,((x-e)*(x-e)*(x-e)-n)/(3*(x-e)*(x-e))
print("Cube root from {} is equal to {:.4f}".format(n,x))
``````

3. By David Ashton

Made by David Ashton. A single line python program to find cube root. Source

`2.1544`
``print(round(int(input())**(1/3), 4))``

4. By Divyanshu Singh

```Enter Number: 10.0

πΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊ
The Cube root of 10.0 is 2.1544
πΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊ```
``````import sys, codecs
sys.stdout = codecs.getwriter('utf_16')(sys.stdout.buffer, 'strict')

def Cube(a):
b = round((a)**(1/3), 4)
return b

a = float(input("Enter Number: "))
print(a)
print()

print('πΊ'*14)
print("The Cube root of", a, "is", Cube(a))
print('πΊ'*14)``````

5. By Victor Maltsev

Made by Victor Maltsev. Python Cube root calculator using Newton method. Source

```Enter a number for calculating the cube root: 12
x(1)=4.666666666666667
x(2)=3.2947845804988667
x(3)=2.564996283933497
x(4)=2.3179736277253107
x(5)=2.2897785669119224
x(6)=2.28942853862757
The cube root of 12.0 is 2.289428485106665```
``````e = 0.0001
try :
y = float(input('Enter a number for calculating the cube root: '))
except :
print('You entered not number!')
else :
i = 0
x = 1
while(True) :
xx = (2*x+y/x/x)/3
d = xx - x
if (abs(d)<e) :
d = y - xx * xx * xx
if (abs(d)<e) :
break
x = xx
i += 1
print('x({0})={1}'.format(i,xx))
print('The cube root of {0} is {1}'.format(y,xx))``````

```Enter the value : 10
2.154434690031884```
``````x=int(input("Enter the value : "))

print(x)

while x>0:
x=x**(1/3)
print(x)
break
``````

7. By Shobhit

```Enter the cube:6.0
The cube root is:1.8171205928321397```
``````n=float(input("Enter the cube:"))
print(n)
x=0.5-(0.5**3-n)/(n*(0.5**2))
for i in range(60000):
x=(lambda x,n:x-(x**3-n)/(n*(x**2)))(x,n)
print(f"The cube root is:{x}")``````

8. By Davy hermans

```7
The cube root of 7.0 is 1.9129```
``````number = float(input())
order = len(str(int(number)))
#print(number)
#print(order)
result = 10**(order/3)
if number < 10:
result = 1
#print(result)
#print(result**3)
def getcloser(r,n):
if abs(r**3 - n) > 0.0001:
if n < 10:
r+= (n-r**3)/10
#            print(r)
getcloser(r,n)
else:
r+= (n-r**3)/n
#            print(r)
getcloser(r,n)
else:
r = round(r,4)
print("The cube root of {0} is {1}".format(n,r))

getcloser(result,number)
``````

9. By Rik Wittkopp

```input is: 9

Cube root is: 2.0797

Test function against input:
8.995018801572998```
``````# Rik Wittkopp

x = int(input())
print('input is:',x)
print()

# Function - find cube root
def root_3():
y = 0
res = y
while res < x:
y += 0.0001
res = round(y**3,2)
if res >= x:
return round(y,4)

print('Cube root is:',root_3())
print()

print('Test function against input:')
print(root_3()**3)

``````

10. By Alan Progg

Made by Alan Progg. Very basic python cube root finder program. Source

```216

216
because
6
x
6
x
6
equals to:
216```
``````num = int(input())
print("")
print("because")
print(num)
print("x")
print(num)
print("x")
print(num)
print("equals to:")
print("")
print("Thanks For Trying")``````

11. By Max

```Number 7
1.9129311827723892```
``````#e is the precision
e=0.0000001
y=float(input("Number\n"))
x=y/2.0
t=x
x=x-(x*x*x*-y)/(3.0*x*x)
#since we do not want our stopping condition to depent on the size of the input we stop if the method stops changing the output greatly
while abs(x-t)>e:
t=x
x=x-(x*x*x-y)/(3.0*x*x)
print(x)
``````