6.138. $\left(x^{2}-6 x\right)^{2}-2(x-3)^{2}=81$.
## Solution.

From the condition $\left(x^{2}-6 x\right)^{2}-2\left(x^{2}-6 x+9\right)-81=0$. Let $x^{2}-6 x=y$.

We have: $y^{2}-2(y+9)-81=0, y^{2}-2 y-99=0 \Rightarrow y_{1}=-9 ; \quad y_{2}=11$. Then 1) $\left.x^{2}-6 x=-9,(x-3)^{2}=0, x_{1,2}=3 ; 2\right) \quad x^{2}-6 x=11, x^{2}-6 x-11=0$, $x_{3,4}=3 \pm \sqrt{20}=3 \pm 2 \sqrt{5}$.

Answer: $x_{1,2}=3 ; x_{3,4}=3 \pm 2 \sqrt{5}$.