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.
 A: $$3a^2+2a+1=0 \to 3a^2+3a+1=a\\3b^2+2b+1=0\to 3b^2+3b+1=b\\a-1=3a(a+1)\\b-1=3b(b+1)$$
so $$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3=\\
\left(\dfrac{-3a(a+1)}{1+a}\right)^3+\left(\dfrac{-3b(b+1)}{1+b}\right)^3\\=-27(a^3+b^3)=-27(s^3-3ps)\\=-27\left(\left(\frac{-2}{3}\right)^3-3\left(\frac{1}{3}\times\left(\frac{-2}{3}\right)\right)\right)\\=+8-18\\=-10$$where $$s=a+b\\p=ab$$
A: 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
$$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3=-10$$
A: The answer is $-10$.
Find the common denominator:
$$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3=\frac{(1-ab-(a-b))^3+(1-ab+(a-b))^3}{(1+ab+(a+b))^3}=$$
$$\frac{2\cdot\left(\frac23\right)^3+2\cdot 3 \cdot\left(\frac23\right)\cdot (a-b)^2}{(\frac{2}{3})^3}=\frac{2\cdot\left(\frac23\right)^3+4\cdot ((a+b)^2-4ab)}{(\frac{2}{3})^3}=\frac{-10(\frac{2}{3})^3}{(\frac23)^3}=-10.$$
A: 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 $2+6\left(\dfrac{2b}{1+b}\right)^2=2+6\cdot(-2)=-10.$
A: 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,
$$x^3+y^3=(x+y)^3-3xy(x+y)$$
$$\left(\frac{1-a}{1+a}\right)^3+\left(\frac{1-b}{1+b}\right)^3=2^3-3\times 2\times3=8-18=-10$$
