Answer:
To simplify the expression \(\frac{a^3b^2(c-3)^2}{(a^2)^2} \div \frac{c}{a}\) and express it as \(\left(\frac{b}{c}\right)^n\), we'll perform the following steps:
1. Simplify the expression inside the parentheses and the numerator.
2. Divide the simplified expression by \(\frac{c}{a}\).
3. Express the result as \(\left(\frac{b}{c}\right)^n\).
Let's proceed step by step:
Given expression:
\[ \frac{a^3b^2(c-3)^2}{(a^2)^2} \div \frac{c}{a} \]
1. Simplify the expression inside the parentheses and the numerator:
\[ \frac{a^3b^2(c^2 - 6c + 9)}{a^4} \]
2. Divide by \(\frac{c}{a}\) is equivalent to multiplying by \(\frac{a}{c}\):
\[ \frac{a^3b^2(c^2 - 6c + 9)}{a^4} \times \frac{a}{c} \]
\[ = \frac{a^4b^2(c^2 - 6c + 9)}{a^4c} \]
\[ = \frac{b^2(c^2 - 6c + 9)}{c} \]
3. Now, we express the result as \(\left(\frac{b}{c}\right)^n\):
\[ \frac{b^2(c^2 - 6c + 9)}{c} = \left(\frac{b}{c}\right)^n \]
To find \(n\), we compare the expression \(\frac{b^2(c^2 - 6c + 9)}{c}\) with \(\left(\frac{b}{c}\right)^n\):
\[ b^2(c^2 - 6c + 9) = b^n c^{n-1} \]
Comparing the exponents and constants on both sides, we see that \(n = 2\), since \(c^{n-1} = c^1 = c\) on the right side.
Therefore, the value of \(n\) is \(2\).
