Given $3x^2+x=1$, find $6x^3-x^2-3x+2010$. 
Given $3x^2+x=1$, find $6x^3-x^2-3x+2010$.

I substituted $3x^2 = 1-x$ in $6x^3-x^2-3x+2010$ and it simplified to
$$\frac{-8x}{3}+2010$$
I know this is a simple problem but I can't solve it. I think there's some method I'm not trying. Please help me with this.

Just synthetic divide and subtract and add till you reduce to something manageable.
$6x^3 -x^2 -3x + 2010 = $
$6x^3+ 2x^2 - 3x^2 - 3x + 2010=$
$2x(3x^3 + x^2) - 3x^2 - 3x+2010=$
$2x - 3x^2 - 3x + 2010=$
$-3x^2 - x + 2010=$
$-(3x^2 + x) +x - x +2010=$
$-1 + 0 + 2010=$
$2009$.
