Problematic polynomial

$\prod { { r }_{ i } } =\frac { 7 }{ 4 }$ by vieta's formulas.So

$\sum _{ k=1 }^{ 6 }{ { { { \left( \frac { \prod { { r }_{ i } } }{ { r }_{ k } } \right) }^{ 4 } } } } ={ \left( \frac { 7 }{ 4 } \right) }^{ 4 }\sum _{ k=1 }^{ 6 }{ \frac { 1 }{ { { r }_{ k } }^{ 4 } } }$

Now to find polynomial whose roots are $\frac { 1 }{ { r }_{ k } }$ for $k=$ $1$ to $6$ is obtained by replacing $x$ with $\frac { 1 }{ x }$ in $f(x)$ .

$f\left( \frac { 1 }{ x } \right) =4\frac { 1 }{ { x }^{ 6 } } +3\frac { 1 }{ { x }^{ 4 } } +2\frac { 1 }{ { x }^{ 3 } } +7=7{ x }^{ 6 }+2{ x }^{ 3 }+3{ x }^{ 2 }=0$

So $f\left( \frac { 1 }{ x } \right)$ has roots $(\frac { 1 }{ { r }_{ k } }$ for $k=$ $1$ to $6$ ).

Now we have to find $\sum _{ k=1 }^{ 6 }{ \frac { 1 }{ { { r }_{ k } }^{ 4 } } }$ .

Now we use newton's sums for $f\left( \frac { 1 }{ x } \right)$ .

We have

$\begin{cases} {{ P }_{ 1 }=0} \\ {{ P }_{ 2 }=0} \\ {{ P }_{ 3 }=\frac { -6 }{ 7 } } \\ { { P }_{ 4 }=\frac { -12 }{ 7 } } \end{cases}$

Here ${ P }_{ n }=\sum _{ i=1 }^{ 6 }{ \frac { 1 }{ { \left( { r }_{ i } \right) }^{ n } } }$

Therefore $\sum _{ k=1 }^{ 6 }{ \frac { 1 }{ { { r }_{ k } }^{ 4 } } } =\frac { -12 }{ 7 }$

Therefore

$\sum _{ k=1 }^{ 6 }{ { { { \left( \frac { \prod { { r }_{ i } } }{ { r }_{ k } } \right) }^{ 4 } } } }$

$={ \left( \frac { 7 }{ 4 } \right) }^{ 4 }\times \frac { -12 }{ 7 }$

$=-16.078$

best way: we get by vietas that $\sum (x^2)=0^2-2*\frac{3}{4}=-\frac{3}{2}$ $\sum (\dfrac{1}{x})=\frac{0}{\frac{7}{4}}=0$ $\prod(x)=\dfrac{7}{4}$ the given expression becomes $\dfrac{-343}{256}\sum (\dfrac{-7}{x^4})$ now, $\begin{array}{c}44x^6+3x^4+2x^3+7=0\\ -7=4x^6+3x^4+2x^3\\ \dfrac{-7}{x^4}=4x^2+3+\frac{2}{x}\\ \sum (\dfrac{-7}{x^4})=4\sum(x^2)+18+2\sum(\frac{1}{x})=4(-\dfrac{3}{2})+18+2*0=12\end{array}$ $\dfrac{-343}{256}\sum (\dfrac{-7}{x^4})=\dfrac{-343}{256}*12=\boxed{-16.078}(3 dp)$

Let $f(x) = \displaystyle \sum_{k=0}^6 a_kx^k = 4x^6+3x^4+2x^3+7$

$\Rightarrow a_0 = 7$ , $a_1 = 0$ , $a_2 = 0$ , $a_3 = 2$ , $a_4 = 3$ , $a_5 = 0$ and $a_6 = 4$ .

$\begin{aligned} \sum_{k=1}^6 \left(\frac{\color{#3D99F6}{\prod r_i}}{r_k} \right)^4 & = \sum_{k=1}^6 \left(\frac{\color{#3D99F6}{ \frac{7}{4}}}{r_k} \right)^4 \quad \quad \quad \small \color{#3D99F6}{\text{By Vieta's formulas } \prod r_i = \frac{a_0}{a_6} = \frac{7}{4}} \\ & = \left(\frac{7}{4}\right)^4 \sum_{k=1}^6 \frac{1}{r_k^4} \quad \quad \small \color{#3D99F6}{\text{By Newton's sums method}} \\ & = \left(\frac{7}{4}\right)^4 \left( \color{#D61F06}{\sum_{i=1}^6 \frac{1}{r_i}} \sum_{k=1}^6 \frac{1}{r_k^3} - \sum_{cyc} \frac{1}{r_ir_j} \sum_{k=1}^6 \frac{1}{r_k^2} + \sum_{cyc} \frac{1}{r_hr_ir_j} \color{#D61F06}{\sum_{k=1}^6 \frac{1}{r_k}} - 4\sum_{cyc} \frac{1}{r_gr_hr_ir_j} \right) \\ & = \left(\frac{7}{4}\right)^4 \left(\color{#D61F06}{\frac{\sum r_fr_gr_hr_ir_j}{\prod r_i}} \sum_{k=1}^6 \frac{1}{r_k^3} - \frac{\sum r_gr_hr_ir_j}{\prod r_i} \sum_{k=1}^6 \frac{1}{r_k^2} + \frac{\sum r_hr_ir_j}{\prod r_i} \color{#D61F06}{\sum_{k=1}^6 \frac{1}{r_k}} - 4\frac{\sum r_ir_j}{\prod r_i} \right) \\ & = \left(\frac{7}{4}\right)^4 \left(\color{#D61F06}{\frac{a_1/a_6}{a_0/a_6}} \sum_{k=1}^6 \frac{1}{r_k^3} - \frac{a_2/a_6}{a_0/a_6} \sum_{k=1}^6 \frac{1}{r_k^2} + \frac{a_3/a_6}{a_0/a_6} \color{#D61F06}{\frac{a_1/a_6}{a_0/a_6}} - 4 \frac{a_4/a_6}{a_0/a_6} \right) \\ & = \left(\frac{7}{4}\right)^4 \left(\frac{a_1}{a_0} \sum_{k=1}^6 \frac{1}{r_k^3} - \frac{a_2}{a_0} \sum_{k=1}^6 \frac{1}{r_k^2} + \frac{a_3}{a_0} \times \frac{a_1}{a_0} - 4\times \frac{a_4}{a_0} \right) \\ & = \left(\frac{7}{4}\right)^4 \left[\frac{0}{4} \sum_{k=1}^6 \frac{1}{r_k^3} - \frac{0}{4} \sum_{k=1}^6 \frac{1}{r_k^2} + \frac{-2}{4}\left(\frac{0}{4}\right) - 4 \left(\frac{3}{4} \right) \right] \\ & = \left(\frac{7}{4}\right)^4 \left[- 4 \left(\frac{3}{4} \right)\right] \\ & = \boxed{-16.078125} \end{aligned}$