Polynomials: Linking equations via their roots The equation $ax^3+bx^2+cx+d=0$ has solutions $1,p$ and $q$. The equation $x^3+sx^2+tx+r=0$ has solutions $1, 1/p$ and $1/q$. Show that $r=a/d$ , and find an expression for $s$ in terms of $c$ and $d$ , simplifying your answer. 

$ax^3+bx^2+cx+d=0\ \ \ \ (1)$
$x^3+sx^2+tx+r=0\  \ \ \ (2)$ 
Clearly the roots of $(1)$ are reciprocal of those of $(2)$
If $y=1/x\iff x=1/y$
Putting the values of $x$ in $(1),$
$a+by+cy^2+dy^3=0\iff dy^3+cy^2+by+a=0\  \ \ \ (3)$ whose roots will be $1,1/p,1/q$
So, $(3)$ will be same as $(2)$
Then the ratios of the coefficients of each powers will be same
