Finding cube root of a number using pow function in C.

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#include<stdio.h>
#include<math.h>
 
main()
{
    float a,b;
     
    printf("Enter a number : ");
    scanf("%d",&a);
     
    b = pow(a,(1/3));
     
    printf("The cube root of a given number is : %f",b);
}

What I have tried:

I have tried using parenthesis. Using 1.0/3.0 , 0.33

Posted 2-May-18 19:30pm

Zeeking99

Updated 2-May-18 23:10pm

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3 solutions

Solution 1

It's pretty simple:

C++

b = pow(a, 1.0 / 3.0);

Or even

C++

b = pow(a, 1 / 3.0);

Posted 2-May-18 19:48pm

OriginalGriff

Updated 2-May-18 19:49pm

v2

Comments

Zeeking99 3-May-18 2:17am

The program is still printing zero.

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CPallini · Accepted Answer · 2018-05-02T23:11:00

Solution 3

Using float variables with pow is pointless. Try

C

#include <stdio.h>
#include <math.h>

int main()
{
  double a, b; // pow takes doubles

  printf("Enter a number : ");

  if ( scanf("%lf",&a) == 1 )
  {
    b = pow(a,(1.0/3));
    printf("The cube root of %lf number is : %lf\n", a, b);
  }

  return 0;
}

Posted 2-May-18 23:11pm

CPallini

Comments

Zeeking99 4-May-18 11:52am

Thanks Man it works. I remembered we can also use implicit casting.
Instead of (1.0/3) we can also use ((float)1/3).
Anyway Thanks for the hint.

Zeeking99 · Accepted Answer · 2018-05-02T20:43:00

I got it. Instead of using 1/3 or 1.0/3.0 you should use 0.333. You will get the right answer.

#include<stdio.h>
#include<math.h>

main()
{
    float a,b;
    
    printf("Enter the number you want to find cube root of : ");
    scanf("%f",&a);
    
    printf("The number you entered is : %.2f\n",a);
    
    b = pow(a,0.333);
    
    printf("The cube root of the number = %.2f",b);
}