According to the fundamental theorem of arithmetic:
$$c = p_1^{m_1} p_2^{m_2} \ldots p_k^{m_k}$$
If we raise both sides by \(n\):
$$ c^n = p_1^{nm_1} p_2^{nm_2} \ldots p_k^{nm_k}$$
If \(ab=c^n\), then some \(p_i\)'s has to be a factor of \(a\) and some \(p_i\)'s has to a factor of \(b\). We can't have a prime that can divide both \(a\) and \(b\) because \((a,b)=1\). This means we can divide the prime factors of \(c^n\) into two groups:
$$ c^n = (\ldots)^n (\ldots)^n$$
The first group contains all the primes that divides \(a\) and the second contains all the groups that divides \(b\). We can use integers \(x\) and \(y\) to represent this:
$$ c^n = x^n y^n$$
If \(c^n=ab\), and \(x^n\) contains all the primes that exactly divide \(a\) while \(y^n\) contains all the primes that exactly divide \(b\), then \(a=x^n\) and \(b=y^n\).