<s>[INST] If the roots of an equation are $a,b,c$ then find the equation having roots $\frac{1-a}{a},\frac{1-b}{b},\frac{1-c}{c}$. Actually I have come to know a technique of solving this kind of problem but it's not exactly producing when used in a certain problem. 
Say we have an equation
$$2x^3+3x^2-x+1=0$$
the roots of this equation are $a,b,c$. If I were to find the equation having the roots $\frac{1}{2a}, \frac{1}{2b}$ and $\frac{1}{2c}$ then I can use the root co-efficient relations and use the rule of creating equations from roots.
This gives me the result : $$4x^3 + 2x^2 -3x -1 =0$$
A much much simpler way to solve this math is: if we pick that, $f(x)= 2x^3+3x^2-x+1$ and the values of $x$ are $a,b$ and $c$ then $f\left(\frac{1}{2x}\right)$ will denote an equation (if we just write $f\left(\frac{1}{2x}\right)=0$ ) which has the roots $\frac{1}{2a}, \frac{1}{2b}$ and $\frac{1}{2c}$. And this way the result also matches the result of my previous work.
But now I face an equation
$$x^3+3x+1=0$$ and if the roots are $a,b,c$ then I have to find the equation with the roots $\frac{1-a}{a},\frac{1-b}{b},\frac{1-c}{c}$.
If I use the root-coefficient relations then I have the result
$$x^3+6x^2+9x+5=0$$ (which is correct I think).
But here $f\left(\frac{1-x}{x}\right)= 3x^3-6x^2+3x-1$. So writing $f\left(\frac{1-x}{x}\right)=0$ won't give me the actual equation. 
Where am I mistaking actually?
 [/INST] The equation $x^3+3x+1=0$ has roots $a,b,c$. Find the equation whose roots are $\frac{1-a}{a},\frac{1-b}{b},\frac{1-c}{c}$.
So let $y=\frac{1-x}{x}$ ... invert this equation, we have $x=\frac{1}{1+y}$, now substitute this into the original equation & you get $\color{red}{y^3+6y^2+9y+5=0}$.
Note that in your previous example you should have made the substitution $y=\frac{1}{2x}$ and then inverted this to $x=\frac{1}{2y}$ ... which of course leads like exactly the same thing.
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