Everyone is playing with Vieta's, so I decided to join in the fun!

Relevant wiki: Vieta's Formula Problem Solving - Intermediate

Call the sum $\mathcal{T}$ . Using Vieta's formula: $\displaystyle\sum_{\text{cyc}}a=7,\displaystyle\sum_{\text{cyc}}ab=5,abc=-2$ $\mathcal T=6\displaystyle \sum_{\text{cyc}} \dfrac{a^3}{\underbrace{abc}_{-2}+a}$ $=6\displaystyle \sum_{\text{cyc}} \dfrac{\overbrace{a^3}^{\color{#D61F06}{8}+(a^3-\color{#D61F06}{8})}}{a-2}$

$=6\displaystyle \sum_{\text{cyc}} \left[\dfrac{8}{a-2}+a^2+2a+4\right]$

$=6\left[ \displaystyle \sum_{\text{cyc}}\dfrac{8}{a-2}+\displaystyle \sum_{\text{cyc}}a^2+2\displaystyle \sum_{\text{cyc}}a+\displaystyle \sum_{\text{cyc}}4\right]$ Now to calculate $\displaystyle \sum_{\text{cyc}}\dfrac{1}{a-2}$ put $x=2$ in $\dfrac{-f'(x)}{f(x)}=\displaystyle \sum_{\text{cyc}}\dfrac{1}{a-x}$ (See my soln here )to obtain $\displaystyle \sum_{\text{cyc}}\dfrac{1}{a-2}=\dfrac{-11}{8}$ while $\displaystyle\sum_{\text{cyc}}a^2=\left(\displaystyle\sum_{\text{cyc}}a\right)^2-2\displaystyle \sum_{\text{cyc}}ab=39$ .

Now , $\mathcal T= 6\left[8\times\dfrac{-11}{8}+39+14+12\right]=\boxed{324}$

Relevant wiki: Vieta's Formula Problem Solving - Intermediate

$\begin{aligned} S & = \sum_{\text{cyc}} \frac {6a^2}{bc+1} \\ & = 6 \sum_{\text{cyc}} \frac {a^2}{\frac {\color{#3D99F6}{abc}}a+1} \quad \quad \small \color{#3D99F6}{\text{Vieta's formulas: }abc = -2} \\ & = 6 \sum_{\text{cyc}} \frac {a^2}{\frac {\color{#3D99F6}{-2}}a+1} \\ & = 6 \sum_{\text{cyc}} \frac {\color{#3D99F6}{a^3}}{a-2} \quad \quad \small \color{#3D99F6}{\text{Since }a^3-7a^2+5a+2=0} \\ & = 6 \sum_{\text{cyc}} \frac {\color{#3D99F6}{7a^2-5a-2}}{a-2} \\ & = 6 \sum_{\text{cyc}} \frac {7(a^2-4a+4)+23(a-2)+16}{a-2} \\ & = 6 \left[7(\color{#3D99F6}{a+b+c})-3(14)+3(23) +16\left(\frac 1{a-2} + \frac 1{b-2} + \frac 1{c-2} \right) \right] \quad \quad \small \color{#3D99F6}{\text{Vieta's: }a+b+c = 7} \\ & = 6 \left[7(\color{#3D99F6}{7})-42+69 +16\left(\frac {\color{#3D99F6}{ab+bc+ca} - 4(a+b+c)+12}{-\color{#D61F06}{(2-a)(2-b)(2-c)}} \right) \right] \quad \quad \small \color{#3D99F6}{\text{Vieta's: }ab+bc+ca = 5} \\ & = 6 \left[76 +16\left(\frac {\color{#3D99F6}{5} - 4(7)+12}{-\color{#D61F06}{(-8)}} \right) \right] \quad \quad \small \color{#D61F06}{(2-a)(2-b)(2-c) = 2^3 -7(2^2) + 5(2)+2 = -8} \\ & = \boxed{324} \end{aligned}$

Relevant wiki: Vieta's Formula Problem Solving - Intermediate

A solution that uses Newton's sums:

From Vieta's formula, we get

$\color{#3D99F6}{a+b+c = 7}\\ \color{#D61F06}{ab+ac+bc = 5}\\ \color{#EC7300}{abc = -2}$

$\displaystyle \sum_{\text{cyc}} \dfrac{6a^2}{bc+1}\\ =\displaystyle \sum_{\text{cyc}} \dfrac{6a^3}{\color{#EC7300}{abc}+a}\\ =\displaystyle \sum_{\text{cyc}} \dfrac{6a^3}{a-2}\\ =\dfrac{6a^3}{a-2}+\dfrac{6b^3}{b-2}+\dfrac{6c^3}{c-2}\\ =\dfrac{6a^3(b-2)(c-2) + 6b^3(a-2)(c-2)+6c^3(a-2)(b-2)}{(a-2)(b-2)(c-2)}\\ =\dfrac{6a^3\big(\color{#EC7300}{bc} - 2(\color{#3D99F6}{b+c})+4\big)+6b^3\big(\color{#EC7300}{ac} - 2(\color{#3D99F6}{a+c})+4\big)+6c^3\big(\color{#EC7300}{ab} - 2(\color{#3D99F6}{a+b})+4\big)}{\color{#EC7300}{abc} -2(\color{#D61F06}{ab+ac+bc})+4(\color{#3D99F6}{a+b+c}) -8}\\ =\dfrac{6a^3\big(\color{#EC7300}{\frac{-2}{a}} - 2(\color{#3D99F6}{7-a})+4\big)+6b^3\big(\color{#EC7300}{\frac{-2}{b}} - 2(\color{#3D99F6}{7-b})+4\big)+6c^3\big(\color{#EC7300}{\frac{-2}{c}} - 2(\color{#3D99F6}{7-c})+4\big)}{\color{#EC7300}{-2} -2(\color{#D61F06}{5})+4(\color{#3D99F6}{7}) -8}\\ =\dfrac{6a^3(\frac{-2}{a}-10+2a)+6b^3(\frac{-2}{b}-10+2b)+6c^3(\frac{-2}{c}-10+2c)}{-2-10+28 -8}\\ =\dfrac{-12a^2-60a^3+12a^4-12b^2-60b^3+12b^4-12c^2-60c^3+12c^4}{8}\\ =\dfrac{12(\color{magenta}{a^4+b^4+c^4})-60(\color{#69047E}{a^3+b^3+c^3})-12(\color{#20A900}{a^2+b^2+c^2})}{8}$

Now, we apply Newton's sums:

$\color{#3D99F6}{a+b+c = 7}\\ \color{#20A900}{a^2+b^2+c^2} = (\color{#3D99F6}{a+b+c})^2 - 2(\color{#D61F06}{ab+ac+bc}) = (\color{#3D99F6}{7})^2 - 2(\color{#D61F06}{5}) = \color{#20A900}{39}\\ \color{#69047E}{a^3+b^3+c^3} = (\color{#3D99F6}{a+b+c})(\color{#20A900}{a^2+b^2+c^2}) - (\color{#D61F06}{ab+ac+bc})(\color{#3D99F6}{a+b+c}) + 3(\color{#EC7300}{abc}) = (\color{#3D99F6}{7})(\color{#20A900}{39}) - (\color{#D61F06}{5})(\color{#3D99F6}{7}) + 3(\color{#EC7300}{-2}) =\color{#69047E}{232}\\ \color{magenta}{a^4+b^4+c^4} = (\color{#3D99F6}{a+b+c})(\color{#69047E}{a^3+b^3+c^3}) - (\color{#D61F06}{ab+ac+bc})(\color{#20A900}{a^2+b^2+c^2}) + \color{#EC7300}{abc}(\color{#3D99F6}{a+b+c}) = (\color{#3D99F6}{7})(\color{#69047E}{232}) - (\color{#D61F06}{5})(\color{#20A900}{39}) + (\color{#EC7300}{-2})(\color{#3D99F6}{7}) =\color{magenta}{1415}$

Substitute these values in:

$\dfrac{12(\color{magenta}{a^4+b^4+c^4})-60(\color{#69047E}{a^3+b^3+c^3})-12(\color{#20A900}{a^2+b^2+c^2})}{8}\\ =\dfrac{12(\color{magenta}{1415})-60(\color{#69047E}{232})-12(\color{#20A900}{39})}{8}\\ =\dfrac{16980-13920-468}{8}\\ =\dfrac{2592}{8}\\ =\boxed{324}$