Solve the equation $8x^3-6x+\sqrt2=0$ The solutions of this equation are given. They are $\frac{\sqrt2}{2}$, $\frac{\sqrt6 -\sqrt2}{4}$ and $-\frac{\sqrt6 +\sqrt2}{4}$. However i'm unable to find them on my own. I believe i must make some form of substitution, but i can't find out what to do. Help would be very appreciated. Thanks in advance.
 A: Using the substitution $x=\frac{t}{\sqrt{2}}$ the equation becomes
$$4{t^3} - 6t + 2 = 0,$$
Using Ruffini you get that
$$4{t^3} - 6t + 2 = \left( {t - 1} \right)\left( {4{t^2} + 4t - 2} \right) = 2\left( {t - 1} \right)\left( {2{t^2} + 2t - 1} \right)$$
This hows that one solution is $t=1$, i.e. $x=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$. To obtain the other two solutions just apply the discriminant formula to the equation ${2{t^2} + 2t - 1}=0$.
A: Let $x=\sqrt{2}\,t$. Substituting, we find that the equation becomes $16\sqrt{2}\,t^3-6\sqrt{2}\,t+\sqrt{2}=0$, which is equivalent to 
$$16t^3-6t+1=0.$$
Now one hopes that the above equation has a rational root. By the Rational Roots Theorem, the possible candidates are $\pm 1$, $\pm\frac{1}{2}$, $\pm\frac{1}{4}$, $\pm \frac{1}{8}$, and $\pm \frac{1}{16}$.
Note that $t=\frac{1}{2}$ works. Now divide the cubic by $t-\frac{1}{2}$ to look for other roots.
Remark: It is a little nicer to note that $t$ is a root of our cubic if and only if $t=\frac{1}{s}$, where $s$ is a root of $s^3-6s^2+16=0$. Then we have the pleasure of not working with fractions. 
A: HINT...Let $x=\cos\theta$
The equation becomes $\cos3\theta=-\frac{1}{\sqrt{2}}$
Can you take it from there?
