And for the denominator, why are there two additional positive
fraction decompositions
Because $1-u^2 = (1-u)(1+u).$ It's the "difference of squares" factorization. And so, squaring both sides, we also have $(1-u^2)^2 = (1-u)^2(1+u)^2$.
In partial fractions decomposition, all irreducible factors must appear as denominators, to the highest power that appears, as well as to all lower powers. So that means we need a denominator for $(1-u)$ and $(1-u)^2,$ but also for $(1+u)$ and $(1+u)^2.$
Where does the $\frac{1}{4}$ come from in the numerator?
For partial fraction decomposition, the numerators must be chosen so that if you were to give every term common denominator and combine the numerators, you would get the starting fraction back. It takes a bit of algebra to find what numerators do the job, but that is where the $\dfrac{1}{4}$ comes from.
But let's do the algebra. The normal way to find these numerators is to introduce variables for them and solve. Like so:
$$
\dfrac{1}{(1-u^2)^2} = \dfrac{1}{(1+u)^2(1-u)^2} = \dfrac{A}{1+u} + \dfrac{B}{(1+u)^2} + \dfrac{C}{1-u} + \dfrac{D}{(1-u)^2}
$$
here the "variable" numerators are called $A, B, C,$ and $D.$
We solve for them by clearing denominators:
$$
1 = A(1-u)^2(1+u) + B(1-u)^2 + C(1-u)(1+u)^2 + D(1+u)^2
$$
Then in general you may have to expand all those binomials and match like terms of the resulting polynomials in $u$. But a shortcut can be to sub in the values of the roots in $u$. For example, at $u=1,$ the above equation becomes
$$
1 = A(1-1)^2(1+1) + B(1-1)^2 + C(1-1)(1+1)^2+ D(1+1)^2 = D(1+1)^2.
$$
So we have $D=1/4.$
Similarly, by next subbing $u=-1,$
$$
1 = B(1-(-1))^2
$$
So we have $B=1/4.$
To figure out that $A$ and $C$ are also $1/4$, we can either sub in some other, non-root values of $u$, or bite the bullet and do the algebra. (But if this were a partial fractions problem with no powers of irreducibles, we'd be done).
I'll do the algebra, I guess. Expand, and gather like terms in $u$, treating the coefficients $A, B, C,$ and $D$ as variables.
$$
1 = A(1-u-u^2+u^3) + B(1-2u+u^2) + C(1+u-u^2-u^3) + D(1+2u+u^2) \\
= (A+B+C+D) + (-A-2B+C+2D)u + (-A+B-C+D)u^2 + (A-C)u^3.
$$
Now for two polynomials in $u$ to be equal for all values of $u$, every coefficient must be equal. On the left-hand side we have the polynomial $1$, which has a constant coefficient of $1$ and all higher coefficients are $0$.
So this gives us the following equations:
$$
1 = A+B+C+D\\
0 = -A-2B+C+2D\\
0 = -A+B-C+D\\
0 = A-C.
$$
We could solve this system of four linear equations in four unknowns, using Gaussian elimination or substitution or matrix methods. But first let's remember that we already know that $B=D=1/4.$ So we don't have four unknowns any more, our only unknowns left are $A$ and $C$. By the fourth equation, $A=C$, so our only unknown is really $A$. The first equation becomes
$$
1 = A+B+C+D = A + \dfrac{1}{4} + A + \dfrac{1}{4}.
$$
So $2A=1-\dfrac{2}{4} = \dfrac{1}{2}.$ And so $A=B=C=D=\dfrac{1}{4}.$
Best Answer
Use the formula twice.
$\begin{align}\cos (x)\cos (2x) \cos(3x) &= \frac12(\cos (4x) + \cos (2x)) \cos (2x) \\&= \frac12\cos(4x)\cos(2x) + \frac12 \cos(2x)\cos(2x) \\&= \frac 14 (\cos(6x) + \cos(2x)) + \frac14(\cos(4x)+1)\end{align}$