Let's telescope

Using $\small{\color{#3D99F6}{x^3+y^3=(x+y)(x^2-xy+y^2)}}$ $P= \displaystyle \prod_{a=3}^{m} \dfrac{(a+2)(a^2-2a+4)}{(a-2)(a^2+2a+4)}$ $=\left(\displaystyle \prod_{a=3}^{m} \dfrac{a+2}{a-2}\right)\times \left(\displaystyle \prod_{a=3}^{m} \dfrac{a^2-2a+4}{a^2+2a+4}\right)$ $=(\dfrac{\not 5}{1}\times \dfrac{\not 6}{2}\times \dfrac{\not 7}{3}\times \dfrac{\not 8}{4}\times\dfrac{\not 9}{\not 5}....\dfrac{\cancel{m+2}}{m-2})\times (\dfrac{7}{\cancel{19}}\times \dfrac{12}{\cancel{28}}\times \dfrac{\cancel{19}}{\cancel{39}}\times...\dfrac{\cancel{m^2-4m+2}}{m^2+4m+2})$ $=\dfrac{84(m-1)m(m+1)(m+2)}{24(m^2+3)(m^2+2m+4)}$ When $m$ $\rightarrow \infty,\color{forestgreen}{P~\text{approaches}~\dfrac{84}{24}}$

Hence, $\Large\color{#302B94}{P=3.5}$

Same method but my presentation is as follows:

$\begin{aligned} P & = \prod_{a=3}^\infty \frac{a^3+8}{a^3-8} \\ & = \prod_{a=3}^\infty \frac{(a+2)(a^2-2a+4)}{(a-2)(a^2+2a+4)} \\ & = \prod_{a=3}^\infty \frac{a+2}{a-2} \prod_{a=3}^\infty \frac{a^2-2a+4}{a^2+2a+4} \\ & = \prod_{a=3}^\infty \frac{a+2}{a-2} \prod_{a=3}^\infty \frac{(a-1)^2+3}{(a+1)^2+3} \\ & = \left(\frac{1}{1\dot{}2\dot{}3\dot{}4} \prod_{a=3}^\infty \frac{a+2}{a+2}\right) \left((2^2+3)(3^2+3)\prod_{a=3}^\infty \frac{(a+1)^2+3}{(a+1)^2+3}\right) \\ & = \frac{84}{24} = \frac{7}{2} = \boxed{3.5} \end{aligned}$

solution lies in the name it is a telescopic multiplicative series a^-8=(a-2)(a^2+2a+4) a^3+8=(a+2)(a^2-2a+4)

$\prod _{ a=3 }^{ \infty }{ \frac { a+2 }{ a-2 } } \prod _{ a=3 }^{ \infty }{ \frac { { a }^{ 2 }-2a+4 }{ { a }^{ 2 }+2a+4 } }$

= 1/24 * 7 * 12 =3.5