Milestone: 100 day streak! (I just want to post a problem)

$\cos A=\dfrac{b^2+c^2-a^2}{2bc}=\left(\dfrac{b^2+c^2-a^2}{2bc}+\color{#D61F06}{1}\right)-\color{#D61F06}{1}$ Now manipulating the first bracket gives:

$\cos A=\left(\dfrac{(b+c)^2-a^2}{2bc}\right)-\color{#D61F06}{1}=\left(\dfrac{(b+c-a)(b+c+a)}{2bc}\right)-\color{#D61F06}{1}$

Now by Vieta's formula: $\color{teal}{\small{\displaystyle\sum_{\text{cyc}}a=24,\displaystyle\sum_{\text{cyc}}ab=180,abc=420}}$ so that:

$\cos A=\dfrac{(24-2a)\cdot 24}{2\cdot\frac{420}a}-\color{#D61F06}{1}=\dfrac{2}{35}\left(12a-a^2\right)-\color{#D61F06}{1}$

Now call the required expression $\mathcal K$ ,

$\begin{aligned}\mathcal K=\displaystyle\sum_{\text{cyc}}\cos A=&\displaystyle\sum_{\text{cyc}}\left[\dfrac{2}{35}(12a-a^2)-\color{#D61F06}{1}\right]\\=&\dfrac{2}{35}[12\cdot 24-((24)^2-2(180))]-\color{#D61F06}{3}\\=&\dfrac{39}{35}\end{aligned}$

$\Large \therefore 11(39+35)+25=\boxed{\color{#007fff}{839}}$

$x^3-24x^2+180x-420=0$

From Vieta's formula , we know that

$\color{#3D99F6}{a+b+c=24}\\ \color{#D61F06}{ab+ac+bc=180}\\ \color{#EC7300}{abc=420}$

From cosine rule , we also know that

$a^2=b^2+c^2-2bc \cos A\\ \implies \cos A = \dfrac{b^2+c^2-a^2}{2bc}$

Similarly, we can find that

$\cos B = \dfrac{a^2+c^2-b^2}{2ac}\\ \cos C = \dfrac{a^2+b^2-c^2}{2ab}$

Therefore,

$\cos A + \cos B + \cos C\\ = \dfrac{b^2+c^2-a^2}{2bc}+ \dfrac{a^2+c^2-b^2}{2ac}+ \dfrac{a^2+b^2-c^2}{2ab}\\ =\dfrac{ab^2+ac^2-a^3+a^2b+bc^2-b^3+a^2c+b^2c-c^3}{2\color{#EC7300}{abc}}\\ =\dfrac{ab(\color{#3D99F6}{a+b})+ac(\color{#3D99F6}{a+c})+bc(\color{#3D99F6}{b+c})-(\color{#69047E}{a^3+b^3+c^3})}{2\color{#EC7300}{abc}}\\ =\dfrac{ab(\color{#3D99F6}{24-c})+ac(\color{#3D99F6}{24-b})+bc(\color{#3D99F6}{24-a})-(\color{#69047E}{a^3+b^3+c^3})}{2\color{#EC7300}{abc}}\\ =\dfrac{24(\color{#D61F06}{ab+ac+bc})-3\color{#EC7300}{abc}-(\color{#69047E}{a^3+b^3+c^3})}{2\color{#EC7300}{abc}}$

Now, we use Newton's sums to get $\color{#69047E}{a^3+b^3+c^3}$ :

$\color{#20A900}{a^2+b^2+c^2}=(\color{#3D99F6}{a+b+c})^2-2(\color{#D61F06}{ab+ac+bc})=(\color{#3D99F6}{24})^2-2(\color{#D61F06}{180})=\color{#20A900}{216}\\ \color{#69047E}{a^3+b^3+c^3} = (\color{#3D99F6}{a+b+c})(\color{#20A900}{a^2+b^2+c^2})-(\color{#D61F06}{ab+ac+bc})(\color{#3D99F6}{a+b+c})+3\color{#EC7300}{abc}= (\color{#3D99F6}{24})(\color{#20A900}{216})-(\color{#D61F06}{180})(\color{#3D99F6}{24})+3(\color{#EC7300}{420})=\color{#69047E}{2124}$

Calculate the sum:

$\cos A + \cos B + \cos C\\ =\dfrac{24(\color{#D61F06}{ab+ac+bc})-3\color{#EC7300}{abc}-(\color{#69047E}{a^3+b^3+c^3})}{2\color{#EC7300}{abc}}\\ =\dfrac{24(\color{#D61F06}{180})-3(\color{#EC7300}{420})-\color{#69047E}{2124}}{2(\color{#EC7300}{420})}\\ =\dfrac{4320-1260-2124}{840}\\ =\dfrac{936}{840}\\ =\dfrac{39}{35}$

$\implies p=39,\;q=35\\ 11(p+q)+25\\ =11(39+35)+25\\ =11(74)+25\\ =814+25\\ =\boxed{839}$