# 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?

• This won't necessarily lead to an answer or give any insight, but have you tried coming up with an example set of roots to see what the correct choice should be? – TomGrubb Nov 20 '17 at 16:57

$P(0)=-1$ 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.