Sure! To solve the expression [tex]\(\left(\frac{a^4 b^2}{a^{-5} b^6}\right)^3\)[/tex], let's break it down step-by-step:
1. Simplify the base inside the parentheses:
The original expression inside the parentheses is [tex]\(\frac{a^4 b^2}{a^{-5} b^6}\)[/tex].
When you divide powers of the same base, you subtract the exponents. So, let's simplify the exponents of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] separately.
For [tex]\(a\)[/tex]:
[tex]\[
a^4 / a^{-5} = a^{4 - (-5)} = a^{4 + 5} = a^9
\][/tex]
For [tex]\(b\)[/tex]:
[tex]\[
b^2 / b^6 = b^{2 - 6} = b^{-4}
\][/tex]
Now, combining these results, we get:
[tex]\[
\frac{a^4 b^2}{a^{-5} b^6} = a^9 b^{-4}
\][/tex]
2. Raise the simplified expression to the power of 3:
Next, we raise [tex]\(a^9 b^{-4}\)[/tex] to the power of 3. When raising a power to another power, you multiply the exponents. So, we get:
For [tex]\(a\)[/tex]:
[tex]\[
(a^9)^3 = a^{9 \cdot 3} = a^{27}
\][/tex]
For [tex]\(b\)[/tex]:
[tex]\[
(b^{-4})^3 = b^{-4 \cdot 3} = b^{-12}
\][/tex]
3. Combine the results:
Therefore, the simplified final expression is:
[tex]\[
\left(\frac{a^4 b^2}{a^{-5} b^6}\right)^3 = a^{27} b^{-12}
\][/tex]
So, we have [tex]\((a^9, b^{-4}, a^{27}, b^{-12})\)[/tex] as the exponents through the steps and the final answer is [tex]\(a^{27} b^{-12}\)[/tex].
