Determining polynomial from roots of another polynomial I am working on an exercize and I know how to more bruteforcely solve it through pure algebra in its simplest form, but it's such a massive mess to demonstrate so I would like to see if there is another way to solve this
Let $x^3 + ax^2+bx+c$ be a given polynomial (with $a,b,c\in Z$ being acctual numbers but for here I am curiuos in the general case) and we know that it has the roots $x_1,x_2,x_3$ which are given, now determine the polynomial that has the following roots $x_1^2$, $x_2^2$ and $x_3^2$
There are several more but I am mostly curious at how one would go about this in a more elegant mannar than just applying the cubical solution formula and then use $p(x)=(x-x_1^2)(x-x_2^2)(x-x_3^2)$ simplifying and such, while possible it is such a hazzle. Is there a more elegant manner?
For the sake of an example, let's take the polynomial $x^3-7x^2+6x-3$ and it has the roots $x_1$ etc, how would I find the coefficients for the new polynomial with the roots of the stated polynomials being squared?

$\!\begin{align} {\rm If}\ \ \color{#c00}{z = x^2}\ \ {\rm then}\ \ 0 = &\ \ \color{#c00}{x^2}x\!+a\color{#c00}{x^2}\!+bx+c\\ = &\ \ \ \color{#c00}z\ x+a\ \color{#c00}z+bx+c\\  \Rightarrow\!&\ \  (z\!+\!b)\,x = -(az\!+\!c)\\ 
\Rightarrow\!&\ \ (z\!+\!b)^2 z =\ \ (az\!+\!c)^2\ \ \text{by squaring prior}\\
\end{align}$ 
The prior equation yields a cubic in $z$ having roots $\, z = x^2$
