What about this? 2

Relevant wiki: Vieta's Formula Problem Solving - Intermediate

Using Vieta's formula: $\displaystyle\sum_{\text{cyc}}\alpha=7,\displaystyle\sum_{\text{cyc}}\alpha\beta=5,\alpha\beta\gamma=-2$

$\mathcal A=\displaystyle\sum_{\text{cyc}}\left(\dfrac{\alpha^4}{\underbrace{\alpha\beta\gamma}_{-2}+\alpha}\right)$

$= \displaystyle\sum_{\text{cyc}}\left(\dfrac{\overbrace{\alpha^4}^{ \color{#EC7300}{16}+(\alpha^4-\color{#EC7300}{16})}}{\alpha-2}\right)$

$=\displaystyle\sum_{\text{cyc}}\left(\dfrac{16}{\alpha-2}+\dfrac{\overbrace{\cancel{(\alpha-2)}(\alpha+2)(\alpha^2+4)}^{\alpha^4-16}}{\cancel{\alpha-2}}\right)$

Now $\small{\displaystyle\sum_{\text{cyc}}\dfrac{1}{\alpha-2}}$ can be calculated using three ways which finally comes out to be $\dfrac{-11}8$ . Now let's calculate second expression i.e $\begin{aligned}&\displaystyle\sum_{\text{cyc}}(\alpha^2+4)(\alpha+2)\\=&\displaystyle\sum_{\text{cyc}}\alpha^3+2\alpha^2+4\alpha+8\\=&\displaystyle\sum_{\text{cyc}}\left(\color{#D61F06}{9\alpha^2}-\color{#20A900}{\alpha}+\color{#3D99F6}{6}\right)~~~\small{\because \alpha^3-7\alpha^2+5\alpha+2=0}\\=&\underbrace{\color{#D61F06}{9\left(\left(\displaystyle\sum_{\text{cyc}}\alpha\right)^2-2\displaystyle\sum_{\text{cyc}}\alpha\beta\right)}}_{\large 9(49-10)=351}-\color{#20A900}{7}+\color{#3D99F6}{6\times 3}\\=&362\end{aligned}$

Hence,

$\mathcal A=16\left(\dfrac{-11}{8}\right)+362$ $\huge =\boxed{340}$

A solution that uses Newton's sums:

For easy typing purposes, let $\alpha = a,\;\beta=b,\;\gamma=c$

From Vieta's formula, we get

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

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

Now, we apply Newton's sums:

$\color{#3D99F6}{a+b+c = 7}\\ \color{#20A900}{a^2+b^2+c^2} = (\color{#3D99F6}{a+b+c})^2 - 2(\color{#D61F06}{ab+ac+bc}) = (\color{#3D99F6}{7})^2 - 2(\color{#D61F06}{5}) = \color{#20A900}{39}\\ \color{#69047E}{a^3+b^3+c^3} = (\color{#3D99F6}{a+b+c})(\color{#20A900}{a^2+b^2+c^2}) - (\color{#D61F06}{ab+ac+bc})(\color{#3D99F6}{a+b+c}) + 3(\color{#EC7300}{abc}) = (\color{#3D99F6}{7})(\color{#20A900}{39}) - (\color{#D61F06}{5})(\color{#3D99F6}{7}) + 3(\color{#EC7300}{-2}) =\color{#69047E}{232}\\ \color{magenta}{a^4+b^4+c^4} = (\color{#3D99F6}{a+b+c})(\color{#69047E}{a^3+b^3+c^3}) - (\color{#D61F06}{ab+ac+bc})(\color{#20A900}{a^2+b^2+c^2}) + \color{#EC7300}{abc}(\color{#3D99F6}{a+b+c})\\ = (\color{#3D99F6}{7})(\color{#69047E}{232}) - (\color{#D61F06}{5})(\color{#20A900}{39}) + (\color{#EC7300}{-2})(\color{#3D99F6}{7}) =\color{magenta}{1415}\\ \color{teal}{a^5+b^5+c^5}= (\color{#3D99F6}{a+b+c})(\color{magenta}{a^4+b^4+c^4}) - (\color{#D61F06}{ab+ac+bc})(\color{#69047E}{a^3+b^3+c^3}) + \color{#EC7300}{abc}(\color{#20A900}{a^2+b^2+c^2})\\ = (\color{#3D99F6}{7})(\color{magenta}{1415}) - (\color{#D61F06}{5})(\color{#69047E}{232}) + (\color{#EC7300}{-2})(\color{#20A900}{39}) =\color{teal}{8667}$

Substitute these values in:

$\mathcal A = \dfrac{2(\color{teal}{a^5+b^5+c^5})-10(\color{magenta}{a^4+b^4+c^4})-2(\color{#69047E}{a^3+b^3+c^3})}{8}\\ =\dfrac{2(\color{teal}{8667})-10(\color{magenta}{1415})-2(\color{#69047E}{232})}{8}\\ =\dfrac{17334-14150-464}{8}\\ =\dfrac{2720}{8}\\ =\boxed{340}$