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How to check whether a number is a perfect cube root in Go

1 Answer

0 votes
// The cube root is a whole number. For example, 27 is a perfect cube, as ∛27 or (27)**1/3 = 3

package main

import (
	"fmt"
	"math"
)

func isPerfectCubeRoot(x int) bool {
	x = int(math.Abs(float64(x)))

	cubeRoot := int(math.Round(math.Pow(float64(x), 1.0/3.0)))

	return math.Pow(float64(cubeRoot), 3) == float64(x)
}

func main() {
	fmt.Println(isPerfectCubeRoot(16))
	fmt.Println(isPerfectCubeRoot(64))
	fmt.Println(isPerfectCubeRoot(27))
	fmt.Println(isPerfectCubeRoot(25))
	fmt.Println(isPerfectCubeRoot(-64))
	fmt.Println(isPerfectCubeRoot(-27))
	fmt.Println(isPerfectCubeRoot(729))
}



/*
run:

false
true
true
false
true
true
true

*/

 



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answered Sep 2 by avibootz
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