The number of points, where the curve $$f(x)=\mathrm{e}^{8 x}-\mathrm{e}^{6 x}-3 \mathrm{e}^{4 x}-\mathrm{e}^{2 x}+1, x \in \mathbb{R}$$ cuts $x$-axis, is equal to _________.

Firstly, we know that the given function <br/><br/>$f(x)=e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1$ intersects the x-axis <br/><br/>where $f(x) = 0$. Setting $f(x)$ equal to zero gives us : <br/><br/>$e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1=0.$ <br/><br/>Let $t = e^{2x}$. The equation now becomes : <br/><br/>$t^4 - t^3 - 3t^2 - t + 1 = 0.$ <br/><br/>Dividing by $t^2$ and rearranging the equation gives : <br/><br/>$t^2 - t - 3 - \frac{1}{t} + \frac{1}{t^2} = 0.$ <br/><br/>If we let $y = t + \frac{1}{t}$, we get : <br/><br/>$y^2 - y - 5 = 0.$ <br/><br/>This quadratic equation in y can be solved using the quadratic formula to give two roots : <br/><br/>$y = \frac{1 \pm \sqrt{21}}{2}.$ <br/><br/>Since $y = t + \frac{1}{t}$, and $t > 0$ (as $t = e^{2x}$), we must choose the root where $y > 2$. Thus, we take $y = \frac{1 + \sqrt{21}}{2}$. <br/><br/>So, we have : <br/><br/>$t + \frac{1}{t} = \frac{1 + \sqrt{21}}{2}.$ <br/><br/>Solving for $t$ gives us : <br/><br/>$t^2 - yt + 1 = 0,$ <br/><br/>or <br/><br/>$t = \frac{y \pm \sqrt{y^2 - 4}}{2}.$ <br/><br/>Substituting $y = \frac{1 + \sqrt{21}}{2}$ into the formula gives : <br/><br/>$t = \frac{\frac{1 + \sqrt{21}}{2} \pm \sqrt{\left(\frac{1 + \sqrt{21}}{2}\right)^2 - 4}}{2}.$ <br/><br/>This quadratic formula for $t$ yields two possible values (t = 2.37 and t = 0.42). Both of them are positive, thus both are acceptable values for $t = e^{2x}$. <br/><br/>Finally, to get $x$, take the natural log of both sides and divide by 2 : <br/><br/>$x = \frac{1}{2}\ln(t).$ <br/><br/>This gives us two solutions for $x$, corresponding to the two positive solutions for $t$. So, the number of points where the curve $f(x) = e^{8x} - e^{6x} - 3e^{4x} - e^{2x} + 1$ intersects the x-axis is 2.