Htykuut

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:
$$abfrac{1}{a}b^2cb^{-3}$$

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?

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
$$
abfrac1abbcfrac1bfrac1bfrac1b
$$
Then we use commutativity to arrange things alphabetically. This gives us
$$
afrac1abbbfrac1bfrac1bfrac1bc
$$
The definition of fraction gives $afrac1a=1$ (and similarly for $b$). We get
$$
1cdot1cdot1cdot1cdot c
$$
And finally, by definition of $1$, this simplifies to $c$.

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.

By commutativity and associativity we can rearrange the expression as
$$left(afrac1aright)cdot(bb^2b^{-3})cdot c$$
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 $c$

All that is needed here from the chapter in question are the associative and commutative properties of multiplication. The expression $abfrac{1}{a}b^2cb^{-3}$ can be re-arranged with commutativity to $afrac{1}{a}bb^2b^{-3}c$, and then this expression can be grouped into $(afrac{1}{a})(bb^2b^{-3})(c)$ 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 $c=5$.