<s>[INST] Why is $ab(\frac1a)b^2cb^{-3} = c$? I'm working through some textbook exercise and am unable to solve the following exercise:

Use the basic rules of algebra to simplify the following expression:
  $$ab\frac{1}{a}b^2cb^{-3}$$

Within the chapter of the book in question the author has covered:


*

*Associative, commutative and distributive properties of an equation

*Expanding brackets

*Factoring

*Quadratic factoring

*Completing the square


The solution in the answers section is simply "$c$". The equation simplifies to $c$. I cannot for the life of me see how. The solution provided does not give the steps in between, only the final solution $c$.
How can one arrive at $c$ with all the steps in between?
 [/INST] $$a \cdot b \cdot \frac{1}{a} \cdot b^2 \cdot c\cdot b^{-3}$$ $$ = a \cdot b\cdot \frac{1}{a} \cdot b^{2} \cdot c\cdot \frac{1}{b^{3}}$$ $$ = a \cdot \frac{1}{a} \cdot b\cdot b^{2} \cdot \frac{1}{b^{3}} \cdot c $$ $$ = \frac{a}{a} \cdot b^{3} \cdot \frac{1}{b^{3}} \cdot c$$ $$ = \frac{a}{a} \cdot \frac{b^{3}}{b^{3}} \cdot c$$$$ = 1 \cdot 1 \cdot c $$$$ = c$$
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