How do you prove $\cos 4x = 8{\cos ^4}x - 8{\cos ^2}x + 1$ ?

Hint: Start this question by twice using the Double angle trigonometric formula for cosine.
expand the formula again and again by substituting other trigonometric formulas.
Expand until it cannot expand further and until we get the desired result.

Formula used: We are going to use some trigonometric formulas:
$\cos 2x = {\cos ^2}x - {\sin ^2}x$
${\sin ^2}x = 1 - {\cos ^2}x$
$\cos 2x = 2{\cos ^2}x - 1$
$\left( {{a^2} - {b^2}} \right) = {a^2} - 2ab + {b^2}$

Complete step-by-step solution:
Firstly, we will expand using the double angle formula for $\cos $.
$\cos 2x = {\cos ^2}x - {\sin ^2}x$
Again, using the formula: ${\sin ^2}x = 1 - {\cos ^2}x$ …. let it be eq. $\left( 1 \right)$
Now, substituting equation $\left( 1 \right)$ in the double angle formula, we get,
$\Rightarrow$$\cos 2x = {\cos ^2}x - {\sin ^2}x$
$\Rightarrow$$\cos 2x = {\cos ^2}x - \left( {1 - } \right.{\cos ^2}\left. x \right)$
On simplifying we get:
$\Rightarrow$$\cos 2x = {\cos ^2}x - 1 + {\cos ^2}x$
$\Rightarrow$$\cos 2x = 2{\cos ^2}x - 1$
Thus, we get: $\cos 2x = 2{\cos ^2}x - 1$ …. let it be eq. $\left( 2 \right)$
So, now to prove $\cos 4x = 8{\cos ^4}x - 8{\cos ^2}x + 1$
We will take L.H.S which can also be written as:
$\Rightarrow$$\cos 4x = \cos \left( {2 \cdot \left( {2x} \right)} \right)$
Let $A = 2x$ …. let it be eq. $\left( 3 \right)$
So, substituting equation $\left( 3 \right)$in the above equation we get
$\Rightarrow$$\cos 4x = \cos \left( {2 \cdot \left( {2x} \right)} \right)$
$\Rightarrow$$\cos 4x = \cos \left( {2 \cdot A} \right)$
$\Rightarrow$$\cos 4x = \cos \left( {2A} \right)$
Substituting equation $\left( 2 \right)$ in the above equation, we get,
$\Rightarrow$$\cos 4x = \cos \left( {2A} \right)$
$\Rightarrow$$\cos 4x = 2{\cos ^2}A - 1$
$\Rightarrow$$\cos 4x = 2{\left( {\cos A} \right)^2} - 1$
Since, $A = 2x$ put it into the above equation and we will get:
$\Rightarrow$$\cos 4x = 2{\left( {\cos 2x} \right)^2} - 1$
Substituting equation $\left( 2 \right)$in the above equation we get:
$\Rightarrow$$\cos 4x = 2{\left( {2{{\cos }^2}x - 1} \right)^2} - 1$
Now, using the formula of $\left( {{a^2} - {b^2}} \right) = {a^2} - 2ab + {b^2}$substituting it in the above equation we get:
$\Rightarrow$$\cos 4x = 2\left( {{{\left[ {2{{\cos }^2}x} \right]}^2} - 2\left[ {2{{\cos }^2}x \cdot 1} \right] + 1} \right) - 1$
Simplifying after expanding the formula:
$\Rightarrow$\[\cos 4x = 2\left( {4{{\cos }^4}x - 4{{\cos }^2}x + 1} \right) - 1\]
$\Rightarrow$\[\cos 4x = 8{\cos ^4}x - 8{\cos ^2}x + 2 - 1\] …. multiplying equation under the bracket with 2.
$\Rightarrow$\[\cos 4x = 8{\cos ^4}x - 8{\cos ^2}x + 1\]
Since the RHS seems like an expansion of the LHS, we will work solely with the LHS to get it to match the RHS.
Summarizing, we have proved: \[\cos 4x = 8{\cos ^4}x - 8{\cos ^2}x + 1\].

Note: Remember that while proving these, you should only change one side of the equation. Pick one side, and try to get it to match the other. I started with LHS as I knew how to expand it and I could foresee that by putting these double angle formulas twice, I will get the RHS.
We should know what formula we used and why. There are three formulas of double angle of cosines- in terms of sines, cosines and tangent. But since both sides of the given equation contain cosines, we're going to use the formula that includes only cosines that is: $\cos 2x = 2{\cos ^2}x - 1$ .