To solve this question, follow the steps below.
Step 01: Factor the denominator of each fraction using (b - 3).
First,
[tex]\begin{gathered} b^2-3b \\ b(b-3) \end{gathered}[/tex]
Second,
[tex]\begin{gathered} b^3-3b^2 \\ b^2(b-3) \end{gathered}[/tex]
Third,
[tex](b-3)[/tex]
So, the expression is:
[tex]\frac{1}{b(b-3)}+\frac{b^2-3b-4}{b^2(b-3)}=\frac{1}{(b-3)}[/tex]
Step 02: Write all fractions together using b²(b-3) as a denominator.
[tex]\frac{b*1+b^2-3b-4=b^2*1}{b^2(b-3)}[/tex]
Solve the multiplications and then the additions/subtractions in the numerator.
[tex]\begin{gathered} \frac{b+b^2-3b-4=b^2}{b^2(b-3)} \\ \frac{b^2-2b-4=b^2}{b^2(b-3)} \end{gathered}[/tex]
Step 03: Remove the denominator.
Since both sides are divided by the same number, you can remove it.
[tex]b^2-2b-4=b^2[/tex]
Step 04: Write all numbers on the same side of the equation.
To do it, subtract b² from both sides.
[tex]\begin{gathered} b^2-2b-4=b^2 \\ b^2-2b-4-b^2=b^2-b^2 \\ -2b-4=0 \end{gathered}[/tex]
Step 05: Solve for b.
To do it, add 4 to both sides and then divide the sides by -2.
[tex]\begin{gathered} -2b-4+4=0+4 \\ -2b=4 \\ \frac{-2b}{-2}=\frac{4}{-2} \\ b=-2 \end{gathered}[/tex]
Answer: b = -2.
