<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] by using the long division
$$\frac{6x^3-x^2-3x+2010}{1}=\frac{6x^3-x^2-3x+2010}{3x^2+x}=(3x^2+x)(2x-1)-\frac{2x-2010}{3x^2+x}$$
$$=(1)(2x-1)-\frac{2x-2010}{1}=2009$$
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