<s>[INST] If $a,b$ are roots of $3x^2+2x+1$ then find the value of an expression It is given that $a,b$ are roots of $3x^2+2x+1$ then find the value of:
$$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3$$
I thought to proceed in this manner:
We know $a+b=\frac{-2}{3}$ and $ab=\frac{1}{3}$. Using this I tried to convert everything to sum and product of roots form, but this way is too complicated! 
Please suggest a simpler process.
 [/INST] Plug $x=\frac{1-y}{1+y}$ in the given equation. We get:
$$\frac{3 (1-y)^2}{(y+1)^2}+\frac{2 (1-y)}{y+1}+1=0$$
Expanding and collecting, we have:
$$y^2-2 y+3=0$$
whose solutions are $$y_1=\frac{1-a}{1+a};\;y_2=\frac{1-b}{1+b}$$
We also know that sum of roots is $s=y_1+y_2=2$ and product is $p=y_1y_2=3$.
The sum of cubes can be written as follows
$$y_1^3+y_2^3=\left(y_1+y_2\right)^3-3y_1y_2(y_1+y_2)=s^3-3ps=8-18=-10$$
so we have
$$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3=-10$$
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