solving $\sqrt{3-\sqrt{3+x}}=x$. Can we solve the following equation in $\mathbb{R}$ without expanding it into a fourth degree equation :
$$ \sqrt{3-\sqrt{3+x}} = x.$$

squaring both sides and squaring again is the only thing I could done, If you have any other idea just post hints.

First, it must be $\,3+x\ge 0\Longleftrightarrow x\ge -3\,$ , for the inner-most square root to be defined on the real field. 
Then it also must be 
$$3-\sqrt{3+x}\ge 0\Longrightarrow9\ge 3+x\Longrightarrow x\le 6$$
for the outer square root to be defined, thus our definition domain is $\,-3\le x \le 6\,$ . Now directly, by successive squaring:
$$x=\sqrt{3-\sqrt{3+x}}\Longrightarrow x^2=3-\sqrt{3+x}\Longrightarrow x^4-6x^2+9=3+x\Longrightarrow$$
$$x^4-6x^2-x+6=0\Longleftrightarrow x^2(x^2-6)-(x-6)=0$$
We get by inspection the zero $\,x=\,$ , so
$$x^4-6x^2-x+6=(x-1)(x^3+x^2-5x-6)$$
Checking for rational solutions to the above we have that $\,x=-2\,$ is a zero, too, so
$$x^4-6x^2-x+6=(x-1)(x+2)(x^2-x-3)$$
Finally, the solutions to that quadratic are
$$x_{1,2}=\frac{1\pm\,\sqrt{13}}{2}=\begin{cases}\frac{1+\sqrt{13}}{2}\cong 2.3\\{}\\\frac{1-\sqrt{13}}{2}\cong -1.3\end{cases}$$
So the roots of the quartic are $\,-2\,,\,-1.3\,,\,1\,,\,2.3\,$ , but since we squared the original irrational equation to get the quartic, we now must check each of the above solutions in the original equation; we get at once that the negative ones must be dropped as we've a square root in one of the sides.
A quick calculator check also rules out the root $\,2.3\,$ of the quartic, so the only real solution is $\,x=1\,$
