What if use pow() with exactly represented degree in approximating step?

    def rootn(x, n):
        g = x**(1.0/n)
        m = 1 << (n-1).bit_length()
        if n != m:
            g = (x*g**(m-n))**(1.0/m)
        return g

Maybe it needs several iterations, because it converges slower than Newton approximation. But every step can be faster.