Correct. The domain of this expression must exclude $\displaystyle{b}=-{1}\usepackage{cmbright}\Huge\$ , -2, -3, and 1, since these values would make either the numerator or the denominator of the divisor equal to 0, and would lead to division by 0. Simplifying the expression, you find $\displaystyle\frac{{{{\left({b}+{2}\right)}}^{{2}}}}{{{b}+{3}}}\cdot\frac{{{\left({b}+{3}\right)}{\left({b}+{1}\right)}}}{{{\left({b}+{2}\right)}{\left({b}-{1}\right)}}}=\frac{{{\left({b}+{2}\right)}{\left({b}+{1}\right)}}}{{{b}-{1}}}\usepackage{cmbright}\Huge\$ .