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$$
Quadratic terms
$$(b+1)x^2=3x^2$$
$$b+1=3$$
$$b=\boxed{2}$$
Linear terms
$$(b-1)x=x$$
$$b-1=1$$
$$b=\boxed{2}$$
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 \(b\).