The roots of
$x^3 + Ax^2 + Bx + C = 0$
are the cubes of the roots of
$x^3 + x^2 - 3x -2.$
Find $A.$

Question

The roots of
$x^3 + Ax^2 + Bx + C = 0$
are the cubes of the roots of
$x^3 + x^2 - 3x -2.$
Find $A.$

anonymous · Answer

If you divide $x^3+x^2-3x-2$ by $x+2$ you get $x^2-x-1$, so one of the roots in -2 and you can use the quadratic formula to find the other two.

Cube all three of them and substitute the cubes into $x^3+Ax^2+Bx+C=0$. The other two roots aren't rational, so it looks like you'll end up with a pretty gnarly system of equations to solve for $A, B,$ and $C$.

$(-2)^3 = -8$, so one of the equations is

$-512+64A-8B+C=0$

thomas5267 · Answer

$$
x^3 + x^2 - 3x -2=0\
(x+2)\left(x^2-x-1\right)=0\
x=-2,\,\frac{1}{2}\left(1+\sqrt{5}\right),\,\frac{1}{2}\left(1-\sqrt{5}\right)\
r_1=\frac{1}{2}\left(1+\sqrt{5}\right)\
r_2=\frac{1}{2}\left(1-\sqrt{5}\right)
$$
$$
\begin{align*}
&\phantom{{}={}}\left(x-r_1^3\right)\left(x-r_2^3\right)\
&=x^2-\left(r_1^3+r_2^3\right)x+(r_1r_2)^3\
&=x^2-\left(\frac{1}{8}\left(1+3\sqrt{5}+15+5\sqrt{5}+1-3\sqrt{5}+15-5\sqrt{5}\right)\right)x-1\
&=x^2-4x-1
\end{align*}
$$
$$
\begin{align*}
&\phantom{{}={}}\left(x^2-4x-1\right)(x+8)\
&=x^3-4x^2-x+8x^2-32x-8\
&=x^3+4x^2-33x-8
\end{align*}\
A=4\
B=-33\
C=-8
$$