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?
 A: Hint:  let $y=(1-x)/x \iff x=1 / (y+1)\,$, then:
$$
P(x)=P\left(\frac{1}{y+1}\right)=\frac{1}{(y+1)^3}\big(1+3(y+1)^2+(y+1)^3\big)
$$
The polynomial in $y$ is therefore:
$$
1+3(y+1)^2+(y+1)^3=y^3 + 6 y^2 + 9 y + 5
$$
This can be verified in WA by calculating the resultant[ x^3+3x+1, 1-x-xy, x ].
A: 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.
A: 
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(1/2x) will denote an equation (if we just write f(1/2x)=0 ) which has the roots 1/2a, 1/2b, 1/2c.

This is not true. Suppose we have $f(x) = (x - 1)^2$. This has the root 1. And $f(1/2x) = (1/2x - 1)^2$. But we solve for the root we get $1/2x - 1 = 0$, so $x = 2 \neq 1/2$.
A: First consider the polynomial with roots $a,b,c$:
$$g(x)=(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc$$
and let 
$$r=a+b+c \quad\quad s=ab+ac+bc \quad\quad t=abc$$
And we accordingly write
$$g(x)=x^3-rx^2+sx-t$$
Now we multiply the linear factors of the expression having the desired roots:
$$f(x)=\Big(x-\frac{1-a}{a}\Big)\Big(x-\frac{1-b}{b}\Big)\Big(x-\frac{1-c}{c}\Big)$$
$$||$$
$$x^3-\frac{ab + ac+bc-3abc}{abc}x^2+\frac{a+b+c+-2(ab+ac+bc)+3abc}{abc}x-\frac{1-(a+b+c)+ab+ac+bc-abc}{abc}$$
and not that each of the coefficients can be expressed with $r,s,t$. Specifically we observe that
$$f(x)=x^3-\frac{s-3t}{t}x^2+\frac{3t-2s+r}{t}x-\frac{1-r+s-t}{t}$$
