Sure, let's break down the given expression step by step:
1. Initial Expression:
[tex]\[
8^{-1}\left(\frac{r y^2}{4 r y^3}\right)^0 \times(8 x)^2
\][/tex]
2. Simplify the middle term:
[tex]\[
\left(\frac{r y^2}{4 r y^3}\right)^0
\][/tex]
Inside the parentheses, observe that both the numerator and the denominator contain the term [tex]\( r \)[/tex]; thus, they cancel each other out:
[tex]\[
\frac{r y^2}{4 r y^3} = \frac{y^2}{4 y^3}
\][/tex]
Now, simplifying [tex]\(\frac{y^2}{4 y^3}\)[/tex]:
[tex]\[
\frac{y^2}{4 y^3} = \frac{1}{4 y}
\][/tex]
Since anything raised to the power of 0 is 1:
[tex]\[
\left(\frac{1}{4 y}\right)^0 = 1
\][/tex]
3. Substitute back into the expression:
[tex]\[
8^{-1} \times 1 \times (8 x)^2
\][/tex]
This simplifies to:
[tex]\[
8^{-1} \times (8 x)^2
\][/tex]
4. Evaluate [tex]\(8^{-1}\)[/tex] and [tex]\((8 x)^2\)[/tex]:
Recall that [tex]\(8^{-1}\)[/tex] is the same as [tex]\(\frac{1}{8}\)[/tex]:
[tex]\[
8^{-1} = \frac{1}{8}
\][/tex]
Now, [tex]\((8 x)^2\)[/tex] means we square both 8 and [tex]\( x \)[/tex]:
[tex]\[
(8 x)^2 = 8^2 \times x^2 = 64 x^2
\][/tex]
5. Combine the terms:
[tex]\[
\frac{1}{8} \times 64 x^2
\][/tex]
Multiplying these together:
[tex]\[
\frac{1}{8} \times 64 x^2 = \frac{64 x^2}{8} = 8 x^2
\][/tex]
6. Conclusion:
The simplified expression is:
[tex]\[
8x^2
\][/tex]
This thorough step-by-step process demonstrates how the problem is simplified and solved. The final answer is [tex]\( 8x^2 \)[/tex].
