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Showing content from https://en.cppreference.com/w/cpp/language/../numeric/math/cbrt.html below:

std::cbrt, std::cbrtf, std::cbrtl - cppreference.com

(1) float       cbrt ( float num );

double      cbrt ( double num );

long double cbrt ( long double num ); (until C++23)

/*floating-point-type*/
            cbrt ( /*floating-point-type*/ num );

(since C++23)
(constexpr since C++26)

float       cbrtf( float num );

(2) (since C++11)
(constexpr since C++26)

long double cbrtl( long double num );

(3) (since C++11)
(constexpr since C++26) template< /*math-floating-point*/ V >

constexpr /*deduced-simd-t*/<V>

            cbrt ( const V& v_num ); (S) (since C++26)

template< class Integer >
double      cbrt ( Integer num );

(A) (constexpr since C++26)

1-3) Computes the cube root of num. The library provides overloads of std::cbrt for all cv-unqualified floating-point types as the type of the parameter.(since C++23)

A) Additional overloads are provided for all integer types, which are treated as double.

(since C++11) [edit] Parameters num - floating-point or integer value [edit] Return value

If no errors occur, the cube root of num (\(\small{\sqrt[3]{num} }\)3√num), is returned.

If a range error occurs due to underflow, the correct result (after rounding) is returned.

[edit] Error handling

Errors are reported as specified in math_errhandling.

If the implementation supports IEEE floating-point arithmetic (IEC 60559),

[edit] Notesstd::cbrt(num)

is not equivalent to

std::pow(num, 1.0 / 3)

because the rational number

\(\small{\frac1{3} }\)

is typically not equal to

1.0 / 3

and

std::pow

cannot raise a negative base to a fractional exponent. Moreover,

std::cbrt(num)

usually gives more accurate results than

std::pow(num, 1.0 / 3)

(see example).

The additional overloads are not required to be provided exactly as (A). They only need to be sufficient to ensure that for their argument num of integer type, std::cbrt(num) has the same effect as std::cbrt(static_cast<double>(num)).

[edit] Example 
#include <cmath>
#include <iomanip>
#include <iostream>
#include <limits>
 
int main()
{
    std::cout
        << "Normal use:\n"
        << "cbrt(729)       = " << std::cbrt(729) << '\n'
        << "cbrt(-0.125)    = " << std::cbrt(-0.125) << '\n'
        << "Special values:\n"
        << "cbrt(-0)        = " << std::cbrt(-0.0) << '\n'
        << "cbrt(+inf)      = " << std::cbrt(INFINITY) << '\n'
        << "Accuracy and comparison with `pow`:\n"
        << std::setprecision(std::numeric_limits<double>::max_digits10)
        << "cbrt(343)       = " << std::cbrt(343) << '\n'
        << "pow(343,1.0/3)  = " << std::pow(343, 1.0 / 3) << '\n'
        << "cbrt(-343)      = " << std::cbrt(-343) << '\n'
        << "pow(-343,1.0/3) = " << std::pow(-343, 1.0 / 3) << '\n';
}

Possible output:

Normal use:
cbrt(729)       = 9
cbrt(-0.125)    = -0.5
Special values:
cbrt(-0)        = -0
cbrt(+inf)      = inf
Accuracy and comparison with `pow`:
cbrt(343)       = 7
pow(343,1.0/3)  = 6.9999999999999991
cbrt(-343)      = -7
pow(-343,1.0/3) = -nan
[edit] See also raises a number to the given power (\(\small{x^y}\)xy)
(function) [edit] computes square root (\(\small{\sqrt{x}}\)√x)
(function) [edit] computes hypotenuse \(\scriptsize{\sqrt{x^2+y^2}}\)√x2
+y2
 and \(\scriptsize{\sqrt{x^2+y^2+z^2}}\)√x2
+y2
+z2
(since C++17)
(function) [edit]

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