Given that $b$ is a constant, what value of $b$ would satisfy the following equation? $$(x^2+bx-1)(x+1)=x^3+3x^2+x-1$$

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Carefully expand the expression on the left side of the equation.

$$(x^2+bx-1)(x+1)$$ $$x^3+bx^2-x+x^2+bx-1$$

Rearranging and combining like terms:

$$x^3+bx^2+x^2+bx-x-1$$ $$x^3+(b+1)x^2+(b-1)x-1$$

Comparing corresponding parts of both equations:

$$x^3+\colorbox{yellow}{\((b+1)x^2\)}+\colorbox{aqua}{\((b-1)x\)}-1 =x^3+\colorbox{yellow}{\(3x^2\)}+\colorbox{aqua}{\(x\)}-1$$

It is not often necessary to expand the entire expression and simplify. If we can identify from the initial expression that there will only be two components that form the quadratic term ($x^2$), we can jump straight to the solution.

$$(x^2+bx-1)(x+1)=x^3+3x^2+x-1$$

The only terms that contribute to the quadratic coefficient are $x^2 \cdot 1$ and $bx \cdot x$ .

Therefore, $bx^2$ and $x^2$ must combine to form $3x^2$.

$b$ must be $\boxed{2}$.

A similar approach can be used for the components that form the linear term to obtain the value of $b$.