Vieta's is fun, so I'll post another problem!

Relevant wiki: Vieta's Formula Problem Solving - Intermediate

*A solution without Newton sums

Using Vieta's : $\small{\displaystyle\sum_{\text{cyc}}a=7,\displaystyle\sum_{\text{cyc}}ab=5,abc=-2}$ $\large \mathcal T=\displaystyle \sum_{\text{cyc}} \dfrac{a^5}{\underbrace{abc}_{-2}+a}=\displaystyle \sum_{\text{cyc}}\dfrac {a^5}{a-2}$

Now, $\small{a^5=(a-2)(a^4+2a^3+4a^2+8a+16)+\color{#20A900}{32}}$ so that: $\mathcal T= \displaystyle \sum_{\text{cyc}}\dfrac{\color{#20A900}{32}}{a-2}+\displaystyle \sum_{\text{cyc}}( \color{#007fff}{a^4}+2\color{magenta}{a^3}+4a^2+8a+16)$

Now, let's manipulate the original equation:

$\color{magenta}{a^3=7a^2-5a-2},~$ multiply both sides by $a$ ,

$\implies \color{#007fff}{a^4}=7\color{magenta}{a^3}-5a^2-2a=\color{#007fff}{44a^2-37a-14}$

Substitute $a^3$ and $a^4$ in second sum of $\mathcal T$ while 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 $\displaystyle \sum_{\text{cyc}}\dfrac{1}{a-2}=\dfrac{-11}{8}$ so that $\small{\displaystyle \sum_{\text{cyc}}\dfrac{\color{#20A900}{32}}{a-2}=-44}$ .

$\large\therefore\mathcal T=-44+ \displaystyle \sum_{\text{cyc}}\left(62a^2-39a-2\right)$

$=-44+\left[62((7)^2-2\times 5)-39(7)-2\times 3\right]$

$\Large =\boxed{\color{#0C6AC7}{2095}}$

Relevant wiki: Newton's Identities

A solution that uses Newton's sums:

From Vieta's formula, we get

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

$\displaystyle \sum_{\text{cyc}} \dfrac{a^4}{bc+1}\\ =\displaystyle \sum_{\text{cyc}} \dfrac{a^5}{\color{#EC7300}{abc}+a}\\ =\displaystyle \sum_{\text{cyc}} \dfrac{a^5}{a-2}\\ =\dfrac{a^5}{a-2}+\dfrac{b^5}{b-2}+\dfrac{c^5}{c-2}\\ =\dfrac{a^5(b-2)(c-2) + b^5(a-2)(c-2)+c^5(a-2)(b-2)}{(a-2)(b-2)(c-2)}\\ =\dfrac{a^5\big(\color{#EC7300}{bc} - 2(\color{#3D99F6}{b+c})+4\big)+b^5\big(\color{#EC7300}{ac} - 2(\color{#3D99F6}{a+c})+4\big)+c^5\big(\color{#EC7300}{ab} - 2(\color{#3D99F6}{a+b})+4\big)}{\color{#EC7300}{abc} -2(\color{#D61F06}{ab+ac+bc})+4(\color{#3D99F6}{a+b+c}) -8}\\ =\dfrac{a^5\big(\color{#EC7300}{\frac{-2}{a}} - 2(\color{#3D99F6}{7-a})+4\big)+b^5\big(\color{#EC7300}{\frac{-2}{b}} - 2(\color{#3D99F6}{7-b})+4\big)+c^5\big(\color{#EC7300}{\frac{-2}{c}} - 2(\color{#3D99F6}{7-c})+4\big)}{\color{#EC7300}{-2} -2(\color{#D61F06}{5})+4(\color{#3D99F6}{7}) -8}\\ =\dfrac{a^5(\frac{-2}{a}-10+2a)+b^5(\frac{-2}{b}-10+2b)+c^5(\frac{-2}{c}-10+2c)}{-2-10+28 -8}\\ =\dfrac{-2a^4-10a^5+2a^6-2b^4-10b^5+2b^6-2c^4-10c^5+2c^6}{8}\\ =\dfrac{2(\color{#624F41}{a^6+b^6+c^6})-10(\color{teal}{a^5+b^5+c^5})-2(\color{magenta}{a^4+b^4+c^4})}{8}$

Now, we apply Newton's sums:

$\color{#3D99F6}{a+b+c = 7}\\ \color{#20A900}{a^2+b^2+c^2} = (\color{#3D99F6}{a+b+c})^2 - 2(\color{#D61F06}{ab+ac+bc}) = (\color{#3D99F6}{7})^2 - 2(\color{#D61F06}{5}) = \color{#20A900}{39}\\ \color{#69047E}{a^3+b^3+c^3} = (\color{#3D99F6}{a+b+c})(\color{#20A900}{a^2+b^2+c^2}) - (\color{#D61F06}{ab+ac+bc})(\color{#3D99F6}{a+b+c}) + 3(\color{#EC7300}{abc}) = (\color{#3D99F6}{7})(\color{#20A900}{39}) - (\color{#D61F06}{5})(\color{#3D99F6}{7}) + 3(\color{#EC7300}{-2}) =\color{#69047E}{232}\\ \color{magenta}{a^4+b^4+c^4} = (\color{#3D99F6}{a+b+c})(\color{#69047E}{a^3+b^3+c^3}) - (\color{#D61F06}{ab+ac+bc})(\color{#20A900}{a^2+b^2+c^2}) + \color{#EC7300}{abc}(\color{#3D99F6}{a+b+c})\\ = (\color{#3D99F6}{7})(\color{#69047E}{232}) - (\color{#D61F06}{5})(\color{#20A900}{39}) + (\color{#EC7300}{-2})(\color{#3D99F6}{7}) =\color{magenta}{1415}\\ \color{teal}{a^5+b^5+c^5}= (\color{#3D99F6}{a+b+c})(\color{magenta}{a^4+b^4+c^4}) - (\color{#D61F06}{ab+ac+bc})(\color{#69047E}{a^3+b^3+c^3}) + \color{#EC7300}{abc}(\color{#20A900}{a^2+b^2+c^2})\\ = (\color{#3D99F6}{7})(\color{magenta}{1415}) - (\color{#D61F06}{5})(\color{#69047E}{232}) + (\color{#EC7300}{-2})(\color{#20A900}{39}) =\color{teal}{8667}\\ \color{#624F41}{a^6+b^6+c^6}= (\color{#3D99F6}{a+b+c})(\color{teal}{a^5+b^5+c^5}) - (\color{#D61F06}{ab+ac+bc})(\color{magenta}{a^4+b^4+c^4}) + \color{#EC7300}{abc}(\color{#69047E}{a^3+b^3+c^3})\\ = (\color{#3D99F6}{7})(\color{teal}{8667}) - (\color{#D61F06}{5})(\color{magenta}{1415}) + (\color{#EC7300}{-2})(\color{#69047E}{232})=\color{#624F41}{53130}$

Substitute these values in:

$\dfrac{2(\color{#624F41}{a^6+b^6+c^6})-10(\color{teal}{a^5+b^5+c^5})-2(\color{magenta}{a^4+b^4+c^4})}{8}\\ =\dfrac{2(\color{#624F41}{53130})-10(\color{teal}{8667})-2(\color{magenta}{1415})}{8}\\ =\dfrac{106260-86670-2830}{8}\\ =\dfrac{16760}{8}\\ =\boxed{2095}$

$Using\ Vieta's \ Formula \ \ for \ f(x)=x^3-7x^2+5x+2=0,\ \ \ \implies\ f(a)=a^3-7a^2+5a+2=0.\\ \displaystyle \ \sum_{cyc}a=7,\ \ \ \sum_{cyc}ab=5,\ \ \ abc=-2, \ \ \ with\ usual\ manipulation\ \sum_{cyc}a=7^2-2*5=39.\\ a \neq 0,\ so\ a^3=7a^2-5a-2,\ \ \ \therefore\ \ a^2=7a-5-\frac 2 a.\\ \implies\ a^3=7*(7a-5-\frac 2 a)\ -5a-2=44a-37- \frac {14} a\ \implies\ \color{#3D99F6}{a^4=44a^2-37a-14}.\\ \displaystyle \sum_{cyc} \dfrac{a^4}{bc+1}= \sum_{cyc} \dfrac{a^5}{abc+a}=\sum_{cyc} \dfrac{a^5}{a-2}=\sum_{cyc} \dfrac{a^5(b-2)(c-2)}{(a-2)(b-2)(c-2)}.\\ (b-2)(c-2)=bc-2(b+c)+4=-\frac 2 a -2(7-a)+4=2a-10-\frac 2 a=\color{#3D99F6}{2(a-5-\frac 1 a)}.\\ a^5(b-2)(c-2)=a^4*a*2(a-5-\frac 1 a)=(44a^2-37a-14)*2*(a^2-5a-1)\\ =2\{44a^4-257a^3-127a^2+107a+14\}\\ = 2\{(44a^2-37a-14)-257(7a^2-5a-2)-127a^2+107a+14\}=2(264a^2 - 236a - 88).\\ (a-2)(b-2)(c-2)=f(2)=8.\\ \therefore \ \displaystyle \sum_{cyc} \dfrac{a^4}{bc+1}=\dfrac{\sum_{cyc}2(264a^2 - 236a - 88)} 8\\ =\sum_{cyc}66\color{#EC7300}{a^2} - 59\color{#EC7300}{a} - 22=66*39-59*7-3*22= \Huge\ \color{#D61F06}{2095}$