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. 

By the fundamental theorem of algebra:
\begin{align}
ax^3+bx^2+cx+d&=a(x-1)(x-p)(x-q)\\&=a\left(x^3-(1+p+q)x^2+(p+q+pq)x-pq\right)\end{align}
and
\begin{align}
x^3+sx^2+tx+r&=(x-1)\left(x-\frac{1}{p}\right)\left(x-\frac{1}{q}\right)\\& = x^3-\left(1+\frac{1}{p}+\frac{1}{q}\right)x^2+\left(\frac{1}{p}+\frac{1}{q}+\frac{1}{pq}\right)-\frac{1}{pq}
\end{align}
Thus $d=-apq$ and $r=-\frac{1}{pq}=\frac{a}{d}$. Also note that $c=a(pq+p+q)$ and $pq=-\frac{d}{a}$, so
$$s=-\left(1+\frac{1}{p}+\frac{1}{q}\right)=-\frac{pq+p+q}{pq}=-\frac{\frac{c}{a}}{-\frac{d}{a}}=\boxed{\frac{c}{d}}.
$$
