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

Question

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

Answer 1

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 $x=y=z$

Answer 2

Like Albus Dumbledore, but a bit of another idea.

By AM-GM $$1=\sum_{cyc}27^{3x^2+2y}\geq3\sqrt[3]{\prod_{cyc}27^{3x^2+2y}}=3^{1+\sum\limits_{cyc}(x^2+2y)}=$$ $$=3^{1+\sum\limits_{cyc}(3x^2+2x)}=3^{\frac{\sum\limits_{cyc}(9x^2+6x+1)}{3}}=3^{\frac{\sum\limits_{cyc}(3x+1)^2}{3}}\geq1,$$ which is possible only for $$x=y=z=-\frac{1}{3}.$$