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

1. anonymous

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 $$-512+64A-8B+C=0$$

2. thomas5267

$x^3 + x^2 - 3x -2=0\\ (x+2)\left(x^2-x-1\right)=0\\ x=-2,\,\frac{1}{2}\left(1+\sqrt{5}\right),\,\frac{1}{2}\left(1-\sqrt{5}\right)\\ r_1=\frac{1}{2}\left(1+\sqrt{5}\right)\\ r_2=\frac{1}{2}\left(1-\sqrt{5}\right)$ \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