Infinite square root muliplication $(x=3\sqrt{y\sqrt{3\sqrt{y}....}})$

Question

I have this problem which says $$ x=3\sqrt{y\sqrt{3\sqrt{y\cdots}}}\\ y=3\sqrt{x\sqrt{3\sqrt{x\cdots}}} $$ What is $x+y$ I tried $$ x^2=9y\sqrt{x}\\ x^3=81y^2\\ y^3=81x^2\\ x^3+y^3=81(x^2+y^2)\\ (x+y)(x^2-xy+y^2)=81(x^2+y^2) $$ I don't know what to do or if I am do that right

Answer 1

Suppose $x \neq 0 \neq y$. From $x^3=81y^2$ and $y^3=81x^2$, it follows that $\frac{x^3}{y^3}=\frac{y^2}{x^2}$, hence $x^5=y^5$, so $x=y$. I think you can take it from there.

Answer 2

From second equality we obtain $$x^2=9x\sqrt{x},$$ which gives $x=0$ or $x=81$. The rest is smooth.