$(x^3y^5z^6)^4 = (xy^2z^3)^7(x^5y^6z^3)$ and we are done.

Fractal's response would require justification as to how he derived it, though it is easily checked. Here is a more rigorous solution: Let $p$ be any factor of $x^3y^5z^6$. Let it's multiplicity in as a factor of $x,y,z$ be $a,b,c$ respectively. Then it's multiplicity in $x^3y^5z^6$ is $3a+5b+6c\equiv 0\mod 7$ Now multiply this congruence by some $k$: $3ka+5kb+6kc\equiv 0\mod 7$ We want $3k\equiv 5\mod 7\ ;\ 5k\equiv 6\mod 7\ ;\ 6k\equiv 3\mod 7$. We can easily find that all of these are satisfied by $k=4$ (we only need to check in the $0\leq k\leq 6$ range). This shows fractal's claim that $(x^3y^5z^6)^4 = (xy^2z^3)^7(x^5y^6z^3)$ and proves that $x^5y^6z^3$ is a $7$th power.

Let $x=p_{1}^{e_1}\cdot \dots p_{k}^{e_k},y=p_{1}^{f_1}\cdot \dots p_{k}^{f_k},z=p_{1}^{g_1}\cdot \dots p_{k}^{g_k}$.Then $1\le \forall i\le k:3e_i+5f_i+6g_i\equiv 0\pmod{7}$.Multiplying congruence by $4$, $5e_i+6f_i+3g_i\equiv 0\pmod{7}$.Therefore $x^5y^6z^3$ is a perfect $7$th power.$\blacksquare$