find a polynomial whose roots are inverse of squares of roots of $x^3+px+q$ Question is :
Given a polynomial $f(x)=x^3+px+q\in \mathbb{Q}[x]$ find a polynomial whose roots are inverse of sqares of roots of $f(x)$
Supposing $a,b,c$ as roots of $f(x)$ we have :


*

*$a+b+c=0$

*$ab+bc+ca=p$

*$abc=-q$


Now i need to know what 


*

*$\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}$

*$\dfrac{1}{a^2}\cdot\dfrac{1}{b^2}+\dfrac{1}{b^2}\cdot\dfrac{1}{c^2}+\dfrac{1}{c^2}\cdot\dfrac{1}{a^2}$

*$\dfrac{1}{(abc)^2}$


All i have to do is use $(a+b+c)^2$ formula and others and conclude what those sums,products are.. I am fairly comfortable with that...
But then, this question was from a Galois theory course.. So, i some how guess there is a better way to do this...
Can some one suggest something..
 A: *

*$x^3+px-q$ has roots $-a,-b,-c$.

*$(x^3+px+q)(x^3+px-q)$ has roots $a,b,c,-a,-b,-c$.  

*$x^3+2px^2+p^2x-q^2$ has roots $a^2,b^2,c^2$.

*$1+2px+p^2x^2-q^2x^3$ has roots $1/a^2,1/b^2,1/c^2$.

A: Hint: start with: $\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} = \dfrac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2} = \dfrac{(ab+bc+ca)^2 - 2abc(a+b+c)}{(abc)^2} = \dfrac{p^2 - 0}{(-q)^2} = \dfrac{p^2}{q^2}$.
A: Note that 
$$
0 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) = a^2 + b^2 + c^2 + 2p
$$
Thus
$$
\dfrac{1}{a^2b^2} + \dfrac{1}{a^2b^2} + \dfrac{1}{a^2b^2} = \dfrac{a^2 + b^2 + c^2}{a^2b^2c^2} = -\dfrac{2p}{q^2}
$$
A: This has been answered by Mr.Michael But then it is not complete answer..
So, Just for my convenience I would write it explicitly... 
Given a polynomial $f(x)\in \mathbb{Q}[x]$ with roots $a,b,c$ we need to find  a poly.  whose roots are $\dfrac{1}{a^2},\dfrac{1}{b^2},\dfrac{1}{c^2}$
Given that $f(x)=x^3+px+q$ with roots $a,b,c$ . So, we have : 


*

*$a+b+c=0$

*$ab+bc+ca=p$

*$abc=-q$


With this we would find polynomial with roots $-a,-b,-c$. For that we need to know :


*

*$(-a)+(-b)+(-c)=-(a+b+c)=0$

*$(-a)(-b)+(-b)(-c)+(-c)(-a)=ab+bc+ca=p$

*$(-a)(-b)(-c)=-abc=q$


So, polynomial with roots $-a,-b,-c$ is $g(x)=x^3+px-q$
So, polynomial with roots $a,b,c,-a,-b,-c$ is  :
$$(x^3+px+q)(x^3+px-q)=x^6+2px^4+p^2x^2-q^2$$
Now we see that $(x-a)(x-b)(x-c)=x^3+px+q$ and $(x+a)(x+b)(x+c)=x^3+px-q$
So, we have $$(x-a)(x-b)(x-c)(x+a)(x+b)(x+c)$$
$$=(x^3+px+q)(x^3+px-q)=x^6+2px^4+p^2x^2-q^2$$
i.e., $$(x-a)(x-b)(x-c)(x+a)(x+b)(x+c)=(x-a)(x+a)(x-b)(x+b)(x-c)(x+c)$$
$$=(x^2-a^2)(x^2-b^2)(x^2-c^2)=x^6+2px^4+p^2x^2-q^2$$
i.e., $$(x^2-a^2)(x^2-b^2)(x^2-c^2)=x^6+2px^4+p^2x^2-q^2$$
So, to get polynomial with roots $a^2,b^2,c^2$ we replace $x^2$ with $y$ 
polynomial with roots $a^2,b^2,c^2$ is $y^3+2py^2+p^2y-q^2$ 
Polynomial with roots $\dfrac{1}{a^2},\dfrac{1}{b^2},\dfrac{1}{c^2}$ is $-q^2x^3+p^2y^2+2py+1$
So, the polynomial that i need is $g(x)=-q^2x^3+p^2y^2+2py+1$
