Applying Combined Exponents Rules: Trinomial

We begin by using the power rule for parentheses:

$(z\cdot t)^n=z^n\cdot t^n$

That is, the power applied to a product inside parentheses, is applied to each of the terms within, when the parentheses are opened.

We then apply the above rule to the problem:

$(2\cdot8\cdot7)^2=2^2\cdot8^2\cdot7^2$

Therefore, the correct answer is option d.

Note:

From the formula of the power property inside parentheses mentioned above, it might seem as though it refers to only two terms of the product inside of the parentheses, but in reality, it is also valid for the power over a multiplication of many terms inside parentheses, as was seen above.

A good exercise is to demonstrate that if the previous property is valid for a power over a product of two terms inside parentheses (as formulated above), then it is also valid for a power over several terms of the product inside parentheses (for example - three terms, etc.).

We use the power law for multiplication within parentheses:

$(x\cdot y)^n=x^n\cdot y^n$

We apply it to the problem:

$(3\cdot4\cdot5)^4=3^4\cdot4^4\cdot5^4$

Therefore, the correct answer is option b.

Note:

From the formula of the power property mentioned above, we understand that it refers not only to two terms of the multiplication within parentheses, but also for multiple terms within parentheses.

We use the power law for multiplication within parentheses:

$$$(x\cdot y)^n=x^n\cdot y^n$

We apply it to the problem:

$(4\cdot7\cdot3)^2=4^2\cdot7^2\cdot3^2$

Therefore, the correct answer is option a.

Note:

From the formula of the power property mentioned above, we understand that we can apply it not only to the multiplication of two terms within parentheses, but is also for multiple terms within parentheses.

We use the formula:

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

$(2^2)^3+(3^3)^4+(9^2)^6=2^{2\times3}+3^{3\times4}+9^{2\times6}=2^6+3^{12}+9^{12}$

To solve the problem$(2\times7\times5)^3$, we need to apply the Power of a Product rule of exponents. This rule states that when you raise a product to a power, it's the same as raising each factor to that power. In mathematical terms, if you have $(abc)^n$, it is equivalent to $a^n \times b^n \times c^n$.

Let's apply this rule step by step:

Our original expression is $(2 \times 7 \times 5)^3$.

We identify the factors inside the parentheses as $2$, $7$, and $5$.

According to the Power of a Product rule, we can distribute the exponent$3$ to each factor:

First, raise $2$ to the power of $3$ to get $2^3$.

Then, raise $7$ to the power of $3$ to get $7^3$.

Finally, raise $5$ to the power of $3$ to get $5^3$.

Therefore, the expression $(2 \times 7 \times 5)^3$ simplifies to $2^3 \times 7^3 \times 5^3$.

Let's handle each expression in the problem separately:

a. We'll start with the leftmost expression, first calculating the result of the multiplication in parentheses, and then use the power rule for power to a power:

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

Let's apply this to the problem for the first expression from the left:

$((7\cdot3)^2)^6=(21^2)^6=21^{2\cdot6}=21^{12}$

where in the final step we calculated the result of multiplication in the power expression,

We're done with this expression, let's move on to the next expression from the left.

b. Let's continue with the second expression from the left, using the power rule for power to a power that we mentioned above and apply it separately to each factor in this expression:

$(3^{-1})^3\cdot(2^3)^4=3^{-1\cdot3}\cdot2^{3\cdot4}=3^{-3}\cdot2^{12}$

Note that the multiplication factors we got have different bases, so we cannot further simplify this expression,

Therefore, let's combine parts a and b above in the result of the original problem:

$((7\cdot3)^2)^6+(3^{-1})^3\cdot(2^3)^4=21^{12}+3^{-3}\cdot2^{12}$

Therefore, the correct answer is answer d.

We use the law of exponents for a power that is applied to parentheses in which terms are multiplied:

$(x\cdot y)^n=x^n\cdot y^n$

We apply the rule to the problem:

$(9\cdot2\cdot5)^3=9^3\cdot2^3\cdot5^3$

When we apply the power within parentheses to the product of the terms we do so separately and maintain the multiplication,

Therefore, the correct answer is option B.