1. **Stating the problem:** Given the equation $$x - \frac{1}{x} = \sqrt{x} + \frac{1}{\sqrt{x}}$$ and the expression $$x^2 + x^{-2}$$, we want to find the value of $$x^2 + x^{-2}$$. 2. **Rewrite the equation:** Let $$a = \sqrt{x}$$, so $$x = a^2$$ and $$\frac{1}{x} = \frac{1}{a^2}$$. The equation becomes: $$a^2 - \frac{1}{a^2} = a + \frac{1}{a}$$ 3. **Introduce new variables:** Let $$y = a - \frac{1}{a}$$ and $$z = a + \frac{1}{a}$$. Note that: $$a^2 - \frac{1}{a^2} = (a - \frac{1}{a})(a + \frac{1}{a}) = yz$$ Given the equation: $$a^2 - \frac{1}{a^2} = a + \frac{1}{a}$$ Substitute: $$yz = z$$ Since $$z \neq 0$$ (because $$a = \sqrt{x} > 0$$), divide both sides by $$z$$: $$y = 1$$ 4. **Find $$x^2 + x^{-2}$$:** Recall: $$x^2 + x^{-2} = a^4 + a^{-4}$$ We can express this in terms of $$z$$: First, note: $$z = a + \frac{1}{a}$$ Square both sides: $$z^2 = a^2 + 2 + \frac{1}{a^2}$$ So: $$a^2 + \frac{1}{a^2} = z^2 - 2$$ Square again to get $$a^4 + a^{-4}$$: $$\left(a^2 + \frac{1}{a^2}\right)^2 = a^4 + 2 + \frac{1}{a^4}$$ Therefore: $$a^4 + \frac{1}{a^4} = \left(a^2 + \frac{1}{a^2}\right)^2 - 2 = (z^2 - 2)^2 - 2$$ 5. **Find $$z$$ using $$y = 1$$:** Recall: $$y = a - \frac{1}{a} = 1$$ Square both sides: $$y^2 = a^2 - 2 + \frac{1}{a^2} = 1$$ So: $$a^2 + \frac{1}{a^2} = y^2 + 2 = 1 + 2 = 3$$ But from step 4: $$a^2 + \frac{1}{a^2} = z^2 - 2$$ Equate: $$z^2 - 2 = 3 \implies z^2 = 5$$ 6. **Calculate $$x^2 + x^{-2}$$:** $$x^2 + x^{-2} = a^4 + a^{-4} = (z^2 - 2)^2 - 2 = (5 - 2)^2 - 2 = 3^2 - 2 = 9 - 2 = 7$$ **Final answer:** $$\boxed{7}$$