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?
 A: First off, associativity means we don't need parentheses, which is good, because there are none given to us. Now, let's use the definition of exponents to get
$$
ab\frac1abbc\frac1b\frac1b\frac1b
$$
Then we use commutativity to arrange things alphabetically. This gives us
$$
a\frac1abbb\frac1b\frac1b\frac1bc
$$
The definition of fraction gives $a\frac1a=1$ (and similarly for $b$). We get
$$
1\cdot1\cdot1\cdot1\cdot c
$$
And finally, by definition of $1$, this simplifies to $c$.
A: First, note that $ab \frac{1}{a} = b$ and then substituting that into the equation we get $bb^2cb^{-3}$ so now we collect all the $b$ terms and the cancel leaving us with $c$ as the simplified expression.
A: By commutativity and associativity we can rearrange the expression as
$$\left(a\frac1a\right)\cdot(bb^2b^{-3})\cdot c$$
the first term is clearly $1$. The second term is also $1$ because of how you add powers in a product: indeed
$$bb^2b^{-3}=b^{1+2-3}=b^0=1$$ So all you are left with is $c$
A: All that is needed here from the chapter in question are the associative and commutative properties of multiplication. The expression $ab\frac{1}{a}b^2cb^{-3}$ can be re-arranged with commutativity to $a\frac{1}{a}bb^2b^{-3}c$, and then this expression can be grouped into $(a\frac{1}{a})(bb^2b^{-3})(c)$ with associativity. Those first two groups are both equal to $1$ (assuming of course none of the variables are $0$), and so you will be left with $c$. If this symbolic manipulation is still confusing, try substituting some values in for those variables! Try $a=2$, $b=3$, and $c=5$.
A: $$a \cdot b \cdot \frac{1}{a} \cdot b^2 \cdot c\cdot b^{-3}$$ $$ = a \cdot b\cdot \frac{1}{a} \cdot b^{2} \cdot c\cdot \frac{1}{b^{3}}$$ $$ = a \cdot \frac{1}{a} \cdot b\cdot b^{2} \cdot \frac{1}{b^{3}} \cdot c $$ $$ = \frac{a}{a} \cdot b^{3} \cdot \frac{1}{b^{3}} \cdot c$$ $$ = \frac{a}{a} \cdot \frac{b^{3}}{b^{3}} \cdot c$$$$ = 1 \cdot 1 \cdot c $$$$ = c$$
A: Assuming that $a$, $b$, and $c$ are real numbers, it is strange that the given expression is claimed equal to $c$. If $a=0$ or $b=0$, this claim is false.
