21i think its correct :D
\hspace{-16}$\textbf{If $\mathbf{x^{10}+(13x+1)^{10}=0}$ has a roots $\mathbf{r_{1}\;,r_{2}\;,\;r_{3}\;,r_{4}\;,r_{5}}$ and $}\\\\ \mathbf{\bar{r_{1}}\;,\bar{r_{2}}\;,\;\bar{r_{3}}\;,\bar{r_{4}}\;,\bar{r_{5}}}.$Then find $\mathbf{\frac{1}{r_{1}.\bar{r_{1}}}+\frac{1}{r_{2}.\bar{r_{2}}}+\frac{1}{r_{3}.\bar{r_{3}}}+\frac{1}{r_{4}.\bar{r_{4}}}+\frac{1}{r_{5}.\bar{r_{5}}}=}$
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7 Answers
1hint please?
in my method im having to calculate product of roots taken 5 at a time ...so im sure im wrong :P
21Here goes the complete proof then,
The equation is \left ( 13+ \frac{1}{x}{} \right )^{10} = 1
or \left ( 13+ \frac{1}{x}{} \right )^{5} = i or \left ( 13+ \frac{1}{x}{} \right )^{5} = -i
The roots of \left ( 13+ \frac{1}{x}{} \right )^{5} = -i will be the conjugate of those that of \left ( 13+ \frac{1}{x}{} \right )^{5} = i
So let the roots of \left ( 13+ \frac{1}{x}{} \right )^{5} = i be r_1, r_2, \cdots ,r_5
From here its easy to get that \frac{1}{r_k}= -13 + cos\left ( \frac{(4k+1)\pi}{10} \right )+ isin\left ( \frac{(4k+1)\pi}{10}\right )
\sum_{k=1}^{5}\frac{1}{|r_k^{2}|}= 850 - 26\sum_{k=1}^{5}\left ( cos\left ( \frac{(4k+1)\pi}{10} \right ) \right )
From the polynomial z^{10} + 1 = 0
conclude immediately that \sum_{k=1}^{5}cos\left ( \frac{{}(4k+1)\pi}{10} \right )= 0
714-5th line kaise hua.. please tell, I'm very weak in everything..
From here its easy" " this part?
62take 5th root on both sides.
RHS will give 5 5th roots of unity
LHS will be 13+xi
take 13 on the other side.
