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.
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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
Made by Divyanshu Singh. Source
Enter Number: 10.0 πΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊ The Cube root of 10.0 is 2.1544 πΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊπΊ
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('πΊ'*14)
print("The Cube root of", a, "is", Cube(a))
print('πΊ'*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))
6. By Anjul Upadhyay
Made by Anjul Upadhyay. Source
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
Made by Shobhit. Source
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
Made by Davy hermans. Source
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
Made by Rik Wittkopp. Source
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 your answer is: 216 because 6 x 6 x 6 equals to: 216
num = int(input())
answer = num*num*num
print(answer)
print("")
print("your answer is:")
print(answer)
print("because")
print(num)
print("x")
print(num)
print("x")
print(num)
print("equals to:")
print(answer)
print("")
print("Thanks For Trying")
11. By Max
Made by Max. Source
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)