A past problem

METHOD-1

The expression is : $\displaystyle\sum_{\text{cyc}} \dfrac{a+2}{a-2}=\displaystyle\sum_{\text{cyc}}\left(1+\dfrac{4}{a-2}\right)$ $=3+ 4\displaystyle\sum_{\text{cyc}}\dfrac 1{a-2}$

$=3+4\left(\dfrac{-f'(2)}{f(2)}\right)=3+4=\boxed 7$

$(\color{#0C6AC7}{\small{\because\ln f(x)= \displaystyle\sum_{\text{cyc}}(x-a)\implies \dfrac{-f'(x)}{f(x)}= \displaystyle\sum_{\text{cyc}}\dfrac{1}{a-x}}})$

METHOD-2

After writing it as $\small{=3+ 4\displaystyle\sum_{\text{cyc}}\dfrac 1{a-2}}$ , use transformation of roots technique s.t $\small{\displaystyle\sum_{\text{cyc}}\dfrac 1{a-2}}$ is the sum of roots of polynomial in $y$ formed by substituting $x=\dfrac{2y+1}{y}$ in original equation. That resulting polynomial is $y^3-y^2\cdots=0$ whose sum is roots is $1$ and hence answer is:

$3+4(1)=\boxed{7}$

METHOD-3

$a^3-4a-8=0\implies \overbrace{4a+8}^{4(a+2)}=a^3\implies a+2=\dfrac{a^3}4$ $a^3-8=4a\implies (a-2)(a^2+2a+4)=4a\implies a-2=\dfrac{4a}{a^2+2a+4}$

Also, $\color{#EC7300}{a^3=4a+8}\implies \color{magenta}{a^4=4a^2+7a}$

Now substituting value of $a-2$ and $a+2$ found earlier, the summation is:

$=\sum_{\text{cyc}}\dfrac{\color{magenta}{a^4}+2\color{#EC7300}{a^3}+4a^2}{16}$ $=\sum_{\text{cyc}}\dfrac{a^2+2a+2}{2}$

$=\dfrac{\left(\left(\displaystyle\sum_{\text{cyc}}a\right)^2-2\displaystyle\sum_{\text{cyc}}ab\right)+2\displaystyle\sum_{\text{cyc}}a+6}{2}$

(By Vieta's formula, $\color{#D61F06}{\small{\displaystyle\sum_{\text{cyc}}a=0,\displaystyle\sum_{\text{cyc}}ab=-4}}$ .)

$=\dfrac{(0-2(-4))+2(0)+6}2=\boxed{7}$

METHOD-4

After writing it as $\small{=3+ 4\displaystyle\sum_{\text{cyc}}\dfrac 1{a-2}}$ , take LCM so that it equals:

$=3+4\times\dfrac{12+(ab+bc+ca)-4(a+b+c)}{abc+4(a+b+c)-2(ab+bc+ca)-8}$

Substituting using Vieta's:

$=3+4\times\dfrac{12+(-4)-0}{8-2(-4)-8}=3+4(1)=\boxed 7$

METHOD-5

After writing it as $\small{=3+ 4\displaystyle\sum_{\text{cyc}}\dfrac 1{a-2}}$ , substitute $\small{a-2=\dfrac{4a}{a^2+2a+4}}$ such that sum is:

$3+\displaystyle\sum_{\text{cyc}}\left(a+2+\dfrac 4a\right)$

$=3+0+6+\displaystyle\sum_{\text{cyc}}\dfrac 4a$

Now $\displaystyle\sum_{\text{cyc}}\dfrac 4a$ can be calculated using LCM , transforming polynomial or simply $\dfrac{-f'(0)}{f(0)}=-2$ . Hence expression is:

$3+0+6-2=\boxed 7$

PS: The solution has already been so long so I would not extend it further ,other wise there are methods left :-)

Method 6:

From Vieta's formula, we know that

$\color{teal}{a+b+c=0}\\ \color{olive}{ab+ac+bc=-4}\\ \color{#624F41}{abc=8}$

Now, to simplify the sum above, we let $\color{#20A900}{a=p+2},\;\color{#69047E}{b=q+2},\;\color{magenta}{c=r+2}$

The sum becomes:

$\dfrac{\color{#20A900}{a}+2}{\color{#20A900}{a}-2}+\dfrac{\color{#69047E}{b}+2}{\color{#69047E}{b}-2}+\dfrac{\color{magenta}{c}+2}{\color{magenta}{c}-2}\\ =\dfrac{\color{#20A900}{p+2}+2}{\color{#20A900}{p+2}-2}+\dfrac{\color{#69047E}{q+2}+2}{\color{#69047E}{q+2}-2}+\dfrac{\color{magenta}{r+2}+2}{\color{magenta}{r+2}-2}\\ =\dfrac{p+4}{p}+\dfrac{q+4}{q}+\dfrac{r+4}{r}\\ =\dfrac{qr(p+4)+pr(q+4)+pq(r+4)}{pqr}\\ =\dfrac{3\color{#EC7300}{pqr}+4(\color{#D61F06}{pq+pr+qr})}{\color{#EC7300}{pqr}}$

Next, we convert the equations as well:

$\color{teal}{a+b+c=0}\\ \color{#20A900}{p+2}+\color{#69047E}{q+2}+\color{magenta}{r+2}=0\\ \implies \color{#3D99F6}{p+q+r=-6}$

$\color{olive}{ab+ac+bc=-4}\\ (\color{#20A900}{p+2})(\color{#69047E}{q+2})+(\color{#20A900}{p+2})(\color{magenta}{r+2})+(\color{#69047E}{q+2})(\color{magenta}{r+2})=-4\\ \color{#D61F06}{pq+pr+qr}+4(\color{#3D99F6}{p+q+r})+12=-4\\ \implies \color{#D61F06}{pq+pr+qr}=-4(\color{#3D99F6}{-6})-12-4=\color{#D61F06}{8}$

$\color{#624F41}{abc=8}\\ (\color{#20A900}{p+2})(\color{#69047E}{q+2})(\color{magenta}{r+2})=8\\ \color{#EC7300}{pqr}+2(\color{#D61F06}{pq+pr+qr})+4(\color{#3D99F6}{p+q+r})+8=8\\ \implies \color{#EC7300}{pqr}=8-2(\color{#D61F06}{8})-4(\color{#3D99F6}{-6})-8=\color{#EC7300}{8}$

Therefore,

$\dfrac{3\color{#EC7300}{pqr}+4(\color{#D61F06}{pq+pr+qr})}{\color{#EC7300}{pqr}}\\ =\dfrac{3(\color{#EC7300}{8})+4(\color{#D61F06}{8})}{\color{#EC7300}{8}}\\ =\dfrac{7(8)}{8}\\ =\boxed{7}$

From Vieta's Formulas, $\begin{cases} {a+b+c= 0} \\ {ab+ac+bc= -4} \\ {abc= 8} \end{cases}$

This is another way of simplifying the algebraic expression given above:

$\dfrac {a+2}{a-2}\ + \dfrac {b+2}{b-2}\ + \dfrac {c+2}{c-2}\\ = \dfrac{a-2}{a-2}\ + \dfrac{4}{a-2}\ + \dfrac{b-2}{b-2}\ + \dfrac{4}{b-2}\ + \dfrac{c-2}{c-2}\ + \dfrac{4}{c-2}\\ = 1 + \dfrac{4}{a-2}\ + 1 + \dfrac{4}{b-2}\ + 1 + \dfrac{4}{c-2}\\ = 3 + 4 \times (\dfrac{1}{a-2}\ + \dfrac{1}{b-2}\ + \dfrac{1}{c-2}\ )\\ = 3 + 4 \times [\dfrac{(b-2)(c-2)+(a-2)(c-2)+(a-2)(b-2)}{(a-2)(b-2)(c-2)}\ ]\\ = 3 + 4 \times [\dfrac{bc-2b-2c+4+ac-2a-2c+4+ab-2a-2b+4}{abc-2ab-2ac+4a-2bc+4b+4c-8}\ ]\\ = 3 + 4 \times [\dfrac{(ab+bc+ac)-4(a+b+c)+12}{abc-2(ab+ac+bc)+4(a+b+c)-8}\ ]\\ = 3 + 4 \times [\dfrac{-4+12}{8-2(-4)+4(0)-8}\ ]\\ = 3 + 4 \times (\dfrac{8}{8}\ )\\ = 3 + 4\\ = \boxed{7}\$

Therefore, $\boxed{7}$ is the numerical value of the expression $\dfrac{a+2}{a-2}\ + \dfrac{b+2}{b-2}\ + \dfrac{c+2}{c-2}$ .