Just an observation

I observed that if we put $x=cosy$ => $y\in [0,\frac{\pi}{2}]$

$2(4cos^{3}y-3cosy)=1$

$cos3y=\frac{1}{2}$

$3y=2n\pi \pm \frac{\pi}{3}$

$y=\frac{2n\pi}{3} \pm \frac{\pi}{9}$

since no value other than $\frac{\pi}{9}$ in $[0,\frac{\pi}{2}]$ is possible so

$x=cos(\frac{\pi}{9})$

$x=0.939$

For the sake of variety (and for practice), we will use the cubic formula.

Let $2x=t$ for simplicity, so that the equation becomes $t^3-3t-1=0$ . The cubic formula gives $t=\sqrt[3]{\frac{1}{2}+\frac{\sqrt{3}}{2}i}+\sqrt[3]{\frac{1}{2}-\frac{\sqrt{3}}{2}i}=\sqrt[3]{e^{\pi{i}/3}}+\sqrt[3]{e^{-\pi{i}/3}}$ $=e^{(1+6k)\pi{i}/9}+e^{-(1+6k)\pi{i}/9}=2\cos((1+6k)\pi/9)$ for $k=0,1,2$ . Only $k=0$ gives a positive solution, $t=2\cos(\frac{\pi}{9})$ , so that $x=\frac{t}{2}=\cos(\frac{\pi}{9})\approx0.940$ .

Assume that X = 1/M Hence, 8(1/M)^3 - 6(1/M) = 1 Hence (8/M^3) - (6/M) = (1/1) Lets make the L.H.S as one part, we have (8M - 6M^3)/M^4 = 1/1 Hence we have the following equations 1 = M^3 and 1 = 8 - 6M^2 >>> 6M^2 = 7 The first equation gives us 3 solutions, two of them are imaginary. ω, ω^2 and ω^3 = 1 none of them are valid lets go to the second equation we have M^2 = 7/6 means that M = 1.08012345 but M = 1/X so X = 0.9258201

Numerical method:

In a calculator let Ans=1

Rearange x=( $\frac {1}{8}+\frac {6Ans}{8}$ ) $^{\frac {1}{3}}$ and type the x value into the calculator

Tap equals until first 3 significant figures of the answer stop changing

x=0.94 (2d.p.)