Don't try to multiply!

$\mathscr R=\left( 1+\cos { \frac { \pi }{ 9 } } \right) \underbrace{\left( 1+\cos { \frac { 3\pi }{ 9 } } \right)}_{\large 1+\left(\frac 12\right)=\frac 32} \left( 1+\cos { \frac { 5\pi }{ 9 } } \right) \left( 1+\cos { \frac { 7\pi }{ 9 } } \right)$ Use $\small{\color{#D61F06}{1+\cos 2x=2\cos^2 x}}$ .

$\mathscr R= 3\left( \color{#20A900}{2\cos \left(\dfrac{\pi}{18} \right)\cos \left(\dfrac{5\pi}{18}\right)}\cos \left(\dfrac{7\pi}{18}\right)\right)^2$

Use $\small{\color{#20A900}{2\cos A\cos B=\cos (A+B)+\cos (A-B)}}$

$\mathscr R=3\left(\left(\color{#20A900}{\underbrace{\cos \left(\dfrac{\pi}3\right)}_{\frac{1}2}+\cos \left(\dfrac{2\pi}9\right)}\right) \cos \left(\dfrac{7\pi}{18}\right)\right)^2$

$=\dfrac 34\left(\cos \left(\dfrac{7\pi}{18}\right)+ 2\cos \left(\dfrac{7\pi}{18}\right)\left(\dfrac{2\pi}9\right)\right)^2$

$=\dfrac 34\left(\cos \left(\dfrac{7\pi}{18}\right)+ \underbrace{\cos \left(\dfrac{11\pi}{18}\right)}_{-\cos \left(\dfrac{7\pi}{18}\right)}+\cos\left(\dfrac{\pi}6\right)\right)^2$

$\large =\dfrac9{16}$

$\Large \therefore 16-9=\huge\boxed{\color{#007fff}{7}}$

$\displaystyle \left ( 1 + \cos{ \dfrac{\pi}{9} } \right )\left ( 1 + \cos{ \dfrac{3\pi}{9} } \right )\left ( 1 + \cos{ \dfrac{5\pi}{9} } \right )\left ( 1 + \cos{ \dfrac{7\pi}{9} } \right )$

Using $\displaystyle \cos{(\pi - x) } = -\cos{x}$ we get:

$\displaystyle \left ( 1 - \cos{ \dfrac{8\pi}{9} } \right )\left ( 1 - \cos{ \dfrac{6\pi}{9} } \right )\left ( 1 - \cos{ \dfrac{4\pi}{9} } \right )\left ( 1 - \cos{ \dfrac{2\pi}{9} } \right )$

Using $\displaystyle 1 - \cos{2x} = 2\sin^2{x}$ we get:

$\displaystyle 2^4 \times \sin^2{\dfrac{4\pi}{9}} \times \sin^2{\dfrac{3\pi}{9}} \times \sin^2{\dfrac{2\pi}{9}} \times \sin^2{\dfrac{\pi}{9}}$

Now using $\displaystyle \sin{(\pi -x)} = \sin{x}$ we get:

$\displaystyle 2^4 \sin{\dfrac{\pi}{9}} \sin{\dfrac{2\pi}{9}} \sin{\dfrac{3\pi}{9}} \sin{\dfrac{4\pi}{9}} \sin{\dfrac{5\pi}{9}} \sin{\dfrac{6\pi}{9}} \sin{\dfrac{7\pi}{9}} \sin{\dfrac{8\pi}{9}}$

Now using the famous identity $\displaystyle \prod_{k=1}^{n-1} \sin{\dfrac{k\pi}{n}} = \dfrac{n}{2^{n-1}}$ we get:

$\displaystyle = 2^4 \times \dfrac{9}{2^8} = \boxed{\dfrac{9}{16}}$

We can use half angle identity and then use the popular identity cosxcos(60-x)cos(60+x) = cos3x / 4