<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] From sum and product, we have:$$a+b+ab+ab=0$$ $$a(b+1)=-b(a+1)$$$$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3=\left(1-\dfrac{2a}{1+a}\right)^3+\left(1-\dfrac{2b}{1+b}\right)^3=\left(1-\dfrac{2ab}{b(1+a)}\right)^3+\left(1-\dfrac{2b}{1+b}\right)^3=\left(1+\dfrac{2b}{1+b}\right)^3+\left(1-\dfrac{2b}{1+b}\right)^3$$
Letting $x=\tfrac{2b}{b+1}$, we have $(1+x)^3+(1-x)^3=2+6x^2=2+6\left(\dfrac{2b}{1+b}\right)^2$.
As, $3b^2+2b+1=0$, so $(b+1)^2=-2b^2$, and finally $2+6\left(\dfrac{2b}{1+b}\right)^2=2+6\cdot(-2)=-10.$
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