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.

Although three years old, this is a good question with some terrific answers. I thought I'd add mine to the collection...
Let$$x=\left(\frac{1-a}{1+a}\right), y=\left(\frac{1-b}{1+b}\right)$$
Then,
$$x+y=\left(\frac{1-a}{1+a}\right)+\left(\frac{1-b}{1+b}\right)=\frac{2(1-ab)}{1+(a+b)+ab}=\frac{2\left(1-\left(\frac{1}{3}\right)\right)}{1+\left(\frac{-2}{3}\right)+\left(\frac{1}{3}\right)}=2$$
and $$xy=\left(\frac{1-a}{1+a}\right)\left(\frac{1-b}{1+b}\right)=\frac{1-(a+b)+ab}{1+(a+b)+ab}=\frac{1-\left(\frac{-2}{3}\right)+\left(\frac{1}{3}\right)}{1+\left(\frac{-2}{3}\right)+\left(\frac{1}{3}\right)}=3$$
From the Binomial Theorem,
$$(x+y)^3=x^3+3x^2y+3xy^2+y^3$$
we get,
$$x^3+y^3=(x+y)^3-3xy(x+y)$$
$$\left(\frac{1-a}{1+a}\right)^3+\left(\frac{1-b}{1+b}\right)^3=2^3-3\times 2\times3=8-18=-10$$
