Elementary Symmetric Polynomials, Roots of cubic polynomials I'm given $a_1, a_2, a_3$ as roots of the equation $x^3 + 7x^2 - 8x + 3$ and need to find the cubic polynomials with roots $a_1^2, a_2^2, a_3^3$ and $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}$.  It's in a chapter about Galois Theory and is preceded by a problem about elementary symmetric polynomials.  I'm guessing I need to use elementary symmetric polynomials to solve for the two desired functions, but I really don't know where to start.  Hints would be much appreciated.  
 A: $a$, $b$ and $c$ are the roots of $x^3 + px^2 + qx + r$ if and only if $abc = -r$, $ab+bc+ca=q$ and $-(a+b+c)=p$.
We have $\frac{1}{a_1} \cdot \frac{1}{a_2}  \cdot\frac{1}{a_3}  = \frac{1}{a_1a_2a_3}  = -1/3$, $\frac{1}{a_1}  + \frac{1}{a_2}  + \frac{1}{a_3}  = \frac{a_1a_2 + a_2a_3 + a_3a_1}{a_1a_2a_3} = \frac{-8}{-3}$ and $\frac{1}{a_1a_2}  + \frac{1}{a_2a_3}  + \frac{1}{a_3a_1}  = \frac{a_1+a_2+a_3}{a_1a_2a_3} = \frac{-7}{-3}$. It follows that $\frac{1}{a_1} $, $\frac{1}{a_2} $ and $\frac{1}{a_3} $ are the roots of $X^3 - \frac{8X^2}{3} + \frac{7X}{3} + \frac{1}{3}$. The problem with $a_1^2$, $a_2^2$, $a_3^2$ can be solved similarly.
A: Extended hints (for a trick that avoids using the elementary symmetric polynomials in the case of squares): 
For any polynomial $p(x)$ we can write the product $p(x)p(-x)$ as polynomial $f(x^2)$ because $p(x)p(-x)$ is even. Furthermore $\deg f=\deg p$. If $\alpha$ is any zero of $p(x)$ show that $\alpha^2$ is a zero of $f(x)$. Here (you do this calculation, too)
$$p(x)p(-x)=9 - 22 x^2 + 65 x^4 - x^6,$$
so we get $f(x)=9-22x+65x^2-x^3.$ Why does this solve your question?
A: Here is another method, which avoids symmetric polynomials - though in the context you might be expected to use them. If fact you can use this method, if you track coefficients, to work out how the symmetric polynomials combine.
For the squares of the roots you need to find a polynomial satisfied by $y=x^2$ and you have $$0=x^3+7x^2-8x+3=x(y-8)+7y+3$$
Whence $$x(8-y)=7y+3$$ Square both sides $$x^2(8-y)^2=(7y+3)^2 \text { or }y(8-y)^2=(7y+3)^2$$
The last equation gives you a cubic in $y$ when you simplify it.

For the second case put $y=\frac 1x$ whence $x=\frac 1y$
$$\frac 1{y^3}+7\frac 1{y^2}-8\frac 1y+3=0$$
Multiply through by $y^3$ to get a cubic in $y$
