Multiplication of Powers: Three Variables

Begin by applying the distributive property of multiplication in order to arrange the algebraic expression according to like terms:

$x^3x^4x^{3+y}y^4z^6$

$$Next, we'll use the law of exponents to multiply terms with the same base:

$a^m\cdot a^n=a^{m+n}$

Note that this law applies to any number of terms being multiplied, not just two. For example, when multiplying three terms with the same base, we obtain the following:

$a^m\cdot a^n\cdot a^k=a^{m+n}\cdot a^k=a^{m+n+k}$

Therefore, we can combine all terms with the same base under one base:

$x^{3+4+3+y}y^4z^6=x^{10+y}y^4z^6$

In the second step we simply added the exponents together.

Note that we could only combine terms with the same base using this law,

From here we can see that the expression cannot be simplified further, and therefore this is the correct and final answer which is answer D.

First, let’s carefully expand the parentheses by applying two exponent laws.

The first law is the product of powers rule, which applies when an exponent is raised over a product of terms:

$(a\cdot b)^n=a^n\cdot b^n$

This law states that when an exponent is applied to parentheses containing a product of terms, the exponent is distributed to each factor inside the parentheses.

The second law we'll use is the power of a power law:

$(a^m)^n=a^{m\cdot n}$

This law states that when an exponent is applied to a term that is already raised to a power (whether written with parentheses for clarity or not), the exponents are multiplied together.

Now, let’s return to the problem and first simplify the two terms with parentheses separately before handling the overall multiplication.

1. The first from left to right is:

$(xyz)^2=x^2\cdot y^2\cdot z^2$

When we expanded the parentheses using the first law, we applied the exponent to each factor in the product inside the parentheses.

2. The second from left to right is:

$(y^x)^y=y^{x y}$

When we expanded the parentheses using the second law, we obtained the following:

$y^3\cdot x^4\cdot(xyz)^2\cdot (y^x)^y=y^3\cdot x^4\cdot x^2\cdot y^2\cdot z^2\cdot y^{xy}$

When we used 1 and 2 mentioned above.

From here on we will no longer indicate the multiplication sign, but use the conventional writing form where placing terms next to each other means multiplication.

Now let's arrange the expression by bases using the multiplication commutative law:

$y^3x^4x^2y^2z^2y^{xy}=y^3y^2y^{xy}x^4x^2z^2$

Let's continue and simplify the expression by using the law of exponents for multiplication between terms with identical bases:

$a^m\cdot a^n=a^{m+n}$

Note that this law is valid for any number of terms in multiplication and not just for two, for example for multiplication of three terms with identical bases we get:

$a^m\cdot a^n\cdot a^k=a^{m+n}\cdot a^k=a^{m+n+k}$

When we used the above exponent law twice, we can also perform the same calculation for four terms in multiplication five and so on..

Let's apply this law to the problem:

$y^3y^2y^{xy}x^4x^2z^2=y^{3+2+xy}x^{4+2}z^2=y^{5+xy}x^6z^2$

When we used the above exponent law for multiplication between terms with identical bases only for terms with the same base.

We got the most simplified expression, so we're done.

Therefore the correct answer is C (when taking into account the commutative law of multiplication).

Important note:

It's worth understanding the reason for the power of a power law mentioned above (the second law), this law comes directly from the definition of exponents:

$(a^m)^n=a^m\cdot a^m\cdot\ldots\cdot a^m=a^{m+m+m+\cdots+m}=a^{m\cdot n}$

When in the first stage we applied the definition of exponents to the term in parentheses and multiplied it by itself n times, then we applied the above law of exponents for multiplication between terms with identical bases and interpreted the multiplication between terms as a sum in the exponent,

Then we used the simple multiplication definition stating that if we connect a number to itself n times we can simply write it as multiplication, meaning:

$m+m+\cdots+m=m\cdot n$

And therefore we get that:

$(a^m)^n=a^{m\cdot n}$

To solve the given problem, we'll apply the laws of exponents to simplify the expression $\frac{a^b b^a}{c^b} \cdot b^{-c} \cdot \frac{1}{a}$.

Let's go through each step:

• Start with the expression $\frac{a^b b^a}{c^b} \cdot b^{-c} \cdot \frac{1}{a}$.
• Rewrite $b^{-c}$ as $\frac{1}{b^c}$ using the rule $x^{-n} = \frac{1}{x^n}$.
• Substitute it back: $\frac{a^b b^a}{c^b} \cdot \frac{1}{b^c} \cdot \frac{1}{a}$.
• Combine the expressions into a single fraction: $\frac{a^b b^a}{c^b \cdot b^c \cdot a}$.
• Use the rule $x^m \cdot x^n = x^{m+n}$ to simplify the exponents in the numerator and denominator: - In the numerator, no changes necessary since terms are already separated. - In the denominator, combine $b^a$ and $b^c$ using the exponent rule: $b^{a+c}$.
• Update the fraction: $\frac{a^b}{a} \cdot \frac{b^a}{b^{a+c}} \cdot \frac{1}{c^b}$.
• Simplify each component: - $\frac{a^b}{a} = a^{b-1}$, - $\frac{b^a}{b^{a+c}} = b^{a-(a+c)} = b^{-c} = \frac{1}{b^c}$.
• Combine all components: $\frac{a^{b-1}}{c^b \cdot b^c}$.
• Express the result combining all simplified terms: $\frac{1}{a^{1-b} b^{c-a} c^b}$.

Therefore, the solution to the problem is $\frac{1}{a^{1-b} b^{c-a} c^b}$.