Simplify $\frac{4\sqrt{7}}{3}\cos{\left(\frac{1}{3}\arccos{\frac{1}{\sqrt{28}}}\right)}+\frac{1}{3}$ If $\dfrac{2\sqrt{19}}{3}\cos{\left(\dfrac{1}{3}\arccos{\dfrac{7}{\sqrt{76}}}\right)}-\dfrac{1}{3}$ can be simpified to $2\left(\cos{\dfrac{4\pi}{19}}+\cos{\dfrac{6\pi}{19}}+\cos{\dfrac{10\pi}{19}}\right)$.
How to simplify  $\dfrac{4\sqrt{7}}{3}\cos{\left(\dfrac{1}{3}\arccos{\dfrac{1}{\sqrt{28}}}\right)}+\dfrac{1}{3}$ ?
edit :
Now, I have get the answer :
$$\dfrac{4\sqrt{7}}{3}\cos{\left(\dfrac{1}{3}\arccos{\dfrac{1}{\sqrt{28}}}\right)}+\dfrac{1}{3}=2\left(\cos{\dfrac{\pi}{7}}+\cos{\dfrac{2\pi}{7}}+\cos{\dfrac{3\pi}{7}}\right)$$
How to prove it?
 A: Call $\theta=\frac{1}{3}\arccos \frac{1}{\sqrt {28}}$. We can use the formula
$\cos (3\theta)=4\cos^3\theta -3\cos \theta$
to deduce
$$\frac{1}{\sqrt {28}}=4\cos^3\theta -3\cos \theta. \tag{1}$$
The question is asking us to evaluate the quantity
$$ y=\frac{4 \sqrt 7}{3}\cos \theta+\frac{1}{3}$$
with $\cos \theta$ satisfying $(1)$. Applying the  substitution $\cos \theta=\frac{3y-1}{4 \sqrt 7}$we get 
$$y^3-y^2-9y=-1. \tag{2}$$ 
The last step is to show that
$$y=2\left(\cos{\dfrac{\pi}{7}}+\cos{\dfrac{2\pi}{7}}+\cos{\dfrac{3\pi}{7}}\right)$$
satisfies $(2)$. This can be done by a straightforward computation, using Werner's formula to deal with the product of cosines. In details we get
$$y^2=2\Big(\cos{\dfrac{2\pi}{7}}+\cos{\dfrac{4\pi}{7}}+2\cos{\dfrac{2\pi}{7}}-\cos{\dfrac{6\pi}{7}}\Big)+6$$
and
$$y^3=y^2\cdot y=24\cos{\frac{\pi}{7}}+24\cos{\frac{2\pi}{7}}+18\cos{\frac{3\pi}{7}}+4\cos{\frac{4\pi}{7}}+2\cos{\frac{5\pi}{7}}+2\cos{\frac{6\pi}{7}}+6.$$
Now we can compute
\begin{align}
y^3-y^2-9y &=2\cos{\frac{\pi}{7}}+4\cos{\frac{2\pi}{7}}+2\cos{\frac{4\pi}{7}}+2\cos{\frac{5\pi}{7}}+4\cos{\frac{6\pi}{7}} \\
&= 2\cos{\frac{2\pi}{7}}-2\cos{\frac{3\pi}{7}}+ 2\cos{\frac{6\pi}{7}}=-1
\end{align}
from which follows
$$y^3-y^2-9y=-1.$$ The last cosines equality is a well known result and it's the only non trivial step.
