What is the general equation of a cubic polynomial? I had this question:
"Find the cubic equation whose roots are the the squares of that of $x^3 + 2x + 1 = 0$"
and I kind of solved it. In that my answer was $x^3 - 4x^2 + 4x + 1$, but it was actually $x^3 + 4x^2 + 4x - 1 = 0$.
I took the general equation of a cubic equation, which was: $x^3 +bx^2/a + cx/a + d/a$.
Through simultaneous equations, I found what $b/a, c/a, d/a$ should equate to for my unknown cubic polynomial. Am I supposed to make $b/a, c/a, d/a$ all positive, then substitute it into the general formula?
Any help would be greatly appreciated, thanks.

Let $a,b,c$ be roots of $x^3 + 2x + 1 = 0$
Using Vieta's Formulas, $a+b+c=0,ab+bc+ca=2,abc=-1$
So, we need to find the cubic equation whose roots are $a^2,b^2,c^2$
i.e., $$(y-a^2)(y-b^2)(y-c^2)=0$$
$$\iff y^3-(a^2+b^2+c^2)y^2+(a^2b^2+b^2c^2+c^2a^2)y+a^2b^2c^2=0$$
Now, $a^2b^2c^2=(abc)^2=(-1)^2=1$
$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=0-2(2)=-4$
$a^2b^2+b^2c^2+c^2a^2=(ab)^2+(bc)^2+(ca)^2$
$=(ab+bc+ca)^2-2(ab\cdot bc+bc\cdot ca+ca\cdot ab)$
$=(ab+bc+ca)^2-2(a+b+c)abc=2^2-2\cdot0\cdot(-1)=4$
