<s>[INST] If roots of $x^6=p(x)$ are given then choose the correct option If $x_1,x_2,x_3,x_4,x_5,x_6$ are real positive roots of the equation $x^6=p(x)$ where $P(x)$ is a 5 degree polynomial where $\frac{x_1}{2}+\frac{x_2}{3}+\frac{x_3}{4}+\frac{x_4}{9}+\frac{x_5}{8}+\frac{x_6}{27}=1$ and $p(0)=-1$, then choose the correct option(s):
$(A)$ $x_5-x_1=x_3x_4$
$(B)$ Product of roots of $P(x)=0$ is $\frac{6}{53}$
$(C)$ $x_2,x_4,x_6$ are in Geometric Progeression
$(D)$ $x_1,x_2,x_3$ are in Arithmetic Progression
Now
$x_1,x_2,x_3,x_4,x_5,x_6$ are real positive roots of the equation $x^6=p(x)$, so I wrote it as 
$x^6-p(x)=(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)(x-x_6)$ but I am not getting how to use the condition $\frac{x_1}{2}+\frac{x_2}{3}+\frac{x_3}{4}+\frac{x_4}{9}+\frac{x_5}{8}+\frac{x_6}{27}=1$ to get the answer. Could someone please help me with this?
 [/INST] $P(0)=-1$ gives $x_1 x_2 x_3 x_4 x_5 x_6 =1$ Now apply AM-GM to $ x_1/2+ x_2/3+ x_3/4+ x_4/9+  x_5/8  + x_6/27 =1$
\begin{eqnarray*}
1 = \frac{x_1}{2 }+ \frac{x_2}{3 }+ \frac{x_3}{ 4}+ \frac{x_4}{9 }+  \frac{x_5}{8 }  + \frac{x_6}{27 } \geq \frac{6 \sqrt[6]{x_1 x_2 x_3 x_4 x_5 x_6}}{6} = 1.
\end{eqnarray*}
For this bound to attained each of the terms in the sum must be $1/6$ so we have 
\begin{eqnarray*}
x_1= \frac{1}{3} ,x_2= \frac{1}{2} ,x_3= \frac{2}{3} ,x_4= \frac{3}{2} ,x_5= \frac{4}{3} ,x_6= \frac{9}{2} . 
\end{eqnarray*}
Quick check $x_5-x_1=1=x_3 x_4$.
$\sum x_i =53/6$ so (B) is also true. ($p(x)=(x_1+ \cdots+x_6)x^5-\cdots-1$).
$x_2,x_4,x_6 = 1/2,3/2,9/2$ are in geometric progression (common ratio $3$).
$x_1,x_2,x_3 = 1/3,1/2,2/3$ are in arithematic progression ( common difference $1/6$).
Thus all $\color{red}{4}$ statements are true.
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