<s>[INST] 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.
 [/INST] $6x^3 = 2x-2x^2$
so 
$6x^3-x^2 -3x+2010 = 2x-2x^2-x^2-3x+2010$
And finally $-x-3x^2+2010 = -x-(1-x)+2010=2009$ 
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