\(\frac{b}{b-3} - \frac{5}{b} = \frac{3}{b-3} \\ \frac{ b^{2} -5(b-3)}{b(b-3)} =\frac{b}{b-3} \\ \frac{ b^{2} -5b+15}{b^2-3b} =\frac{b}{b-3} \\ (b-3)(b^{2} -5b+15)=b(b^2-3b) \\ b^3-8b^2+30b-45=b^3-3b^2 \\ 5b^2-30b+45=0 \\ b^2-6b+9=0 \\ (b -3)(b-3)=0 \\ b=3\)
To check: 3/3-3 - 5/3 = 3/0 - 5/3 which is not defined.
Therefore, b = 3 is an extraneous solution.
