Vieta's might come in handy (Polynomials Sprint question)

Clever solution:

Note that $6x^3-3x^2+2x-1=6(x-\alpha)(x-\beta)(x-\gamma)$

Also, $1+\alpha+\beta+\gamma+\alpha\beta+\beta\gamma+\gamma\alpha+\alpha\beta\gamma=(\alpha+1)(\beta+1)(\gamma+1)$

Now plug in $x=-1$ in our first equation to get that $6(-1-\alpha)(-1-\beta)(-1-\gamma)=6(-1)^3-3(-1)^2+2(-1)-1=-12$ or $-(1+\alpha)(1+\beta)(1+\gamma)=-2$

Thus, $(1+\alpha)(1+\beta)(1+\gamma)=\boxed{2}$

$6x^3-3x^2+2x-1=0 \implies x^3-\frac{1}{2}x^2+\frac{1}{3}x-\frac{1}{6}=0$

$(x-\alpha)(x-\beta)(x-\gamma)=x^3-(\alpha+\beta+\gamma)x^2+(\alpha \beta + \beta \gamma + \alpha \gamma)x-(\alpha \beta \gamma)$

$1+\alpha+\beta+\gamma+\alpha \beta + \beta \gamma + \alpha \gamma+\alpha \beta \gamma = 1 + \frac{1}{2} +\frac{1}{3}+\frac{1}{6}= \boxed{2}$

1 + ( Sum of roots) + (Sum of roots two at a time) + (Products of roots).......

And we get the results......... 1 + (1/2) + (1/3) + (1/6)

$6x^3-3x^2+2x-1=3x^2(2x-1)+(2x-1)=(2x-1)(3x^2+1)$
Now, it's easy to find the roots of this polynomial in $\mathbb{C}$ that are: $\alpha=1/2 ,\hspace{1cm} \beta = i/\sqrt{3} \hspace{1cm}$ and $\hspace{1cm} \gamma= -i/\sqrt{3}$
After calculating, we find that the answer is $2$ .