Let $a$ be a root of $x^3+2x+1=0$ and $b=a^2$ be a root of the required equation
So, $a^3+2a+1=0\implies a\cdot b+2a+1=0\implies a=-\frac1{b+2}$
As $a$ be a root of $x^3+2x+1=0$, put this value of $a$ in $x^3+2x+1=0$
On simplification, I get $(b+2)^3-2(b+2)^2-1=0\iff b^3+4b^2+4b-1=0$
So, the required equation will be $y^3+4y^2+4y-1=0$
I'm going to post the method involving trig functions. As an example, we will be using $0=x^3-3x^2+3$.
To use this method, you must know that
$$\cos(3\theta)=4\cos^3(\theta)-3\cos(\theta)$$
which is easily derived from the sum of angles formulas and the Pythagorean theorem.
We manipulate this to give us
$$\cos(3\arccos(r))=4r^3-3r\tag0$$
Anyways, we start with
$$0=ax^3+bx^2+cx+d$$
$$0=x^3-3x^2+3$$
We always start with the substitution $x=y-\frac{b}{3a}$, which for our example is $x=y+1$
$$0=(y+1)^3-3(y+1)^2+3\\=y^3-3y+1$$
In general, we get something along the lines of
$$0=a\left(y-\frac{b}{3a}\right)^3+b\left(y-\frac{b}{3a}\right)^2+c\left(y-\frac{b}{3a}\right)+d\\=ay^3+py+q$$
Where $p$ and $q$ are constants.
Make the substitution $y=uz$ and multiply both sides by $v$.
$$0=v(uz)^3-3v(uz)+3v\\=vu^3z^3-3vuz+v$$
And have $vu^3=4$ and $-3vu=-3$ so that it comes in our $(0)$ form. Solving this system of equations gives $u=2$ and $v=\frac12$.
$$0=4z^3-3z+\frac12$$
$$0=\cos(3\arccos(z))+\frac12$$
Solving for $z$ and recalling the period of cosine, we get
$$z=\cos\left(\frac{2\pi k}3+\frac13\arccos\left(\frac{-1}2\right)\right)$$
$$y=2z=2\cos\left(\frac{2\pi k}3+\frac13\arccos\left(\frac{-1}2\right)\right)$$
$$x=1+y=1+2\cos\left(\frac{2\pi k}3+\frac13\arccos\left(\frac{-1}2\right)\right)$$
$$x=1+2\cos\left(\frac{2\pi k}3+\frac{2\pi}9\right)\qquad k=1,2,3$$
This is not always possible, but under those cases, we can just switch to different trig functions and use their corresponding formulas. It's pretty easy to see which trig function you should use at the "solve for $u$ and $v$ step".
An advantage to this method is that it avoids casus irreducibilis, which is a fancy way of saying you can't factor a cubic polynomial using only real numbers, radicals, and basic arithmetic operations. This does not include trig functions, however, which can be used to obtain nice forms of the solution.
To compare, the radical solution for the above polynomial is given as
$$x_1=1-\frac12(1-i\sqrt3)\sqrt[3]{\frac12(-1+i\sqrt3)}-\frac{1+i\sqrt3}{\sqrt[3]{4(-1+i\sqrt3}}$$
$$x_2=1-\frac{1-i\sqrt3}{\sqrt[3]{4(-1+i\sqrt3)}}-\frac12(1+i\sqrt3)\sqrt[3]{\frac12(-1+i\sqrt3)}$$
$$x_3=1+\frac1{\sqrt[3]{\frac12(-1+i\sqrt3)}}+\sqrt[3]{\frac12(-1+i\sqrt3)}$$
Compared side by side, you might find one form much nicer.
Best Answer
Suppose you have a root $x$ of $$3x^3-2x^2+1=0$$
If $y=2x$, then $x=\cfrac y 2$ and $$3(\frac y 2)^3-2(\frac y 2)^2+1=0$$
Clear the fractions and you obtain an equation with integer coefficients satisfied by $y$.
The key to the substitution method is to find an expression for $x$ (a root of the original equation) and substitute it into the original equation. It depends on doing the same thing to each of the roots.