find all $(x,y,z)$ such that $27^{3x^2 + 2y}+27^{3y^2 + 2z}+27^{3z^2 + 2x}=1$ 
Find all ($x,y,z$) such that
$$ 27^{3x^2 + 2y}+27^{3y^2 + 2z}+27^{3z^2 + 2x}=1$$

I am a high school student and would appreciate it if anyone could solve it using high school mathematics.
here is what I have tried till now:
$$ 3x^2 +2y < 0 $$ (this can also be said for the other 3 equation. it also means that x,y,z<0)
$$ -2y>3x^2 $$
$$ 4y^2>9x^2 $$ ( -2y and 3x^2 are both positive)
$$ 3y^2>27/4 x^2 $$
$$ -2z>3y^2>27/4 x^4 $$
$$ 4z^2 > 729/16 x^8 $$
$$ 3z^2> 2187/64 x^8 $$
$$ -128/2187<x^7 $$
$$ -2/3 < x $$
$$ -2/3< x,y,z < 0 $$
i tried this but as you can see didn't do much good

Hint:By am-gm $$1\ge3\sqrt[3]{{27}^{3(x^2+y^2+z^2)+2(x+y+z)}} = 3^{3(x^2+y^2+z^2)+2(x+y+z)+1}..(1)$$
But $$3(x^2+y^2+z^2)\ge {(x+y+z)}^2$$ when x=y=z
substituting in  (1)
$$1\ge 3^{{(x+y+z+1)}^2}$$
which is possible only when $x+y+z=-1$ and $x=y=z$
