Question

simplify.
\\(\left(\frac{b^2c^2}{3b^4}\
ight)^3\\)
\\(\frac{c^{?}}{\square b^{\square}}\\)

Explanation:

Step 1: Simplify the numerator and denominator inside the parentheses

First, simplify \(\frac{b^{2}c^{2}}{3b^{4}}\). Using the rule of exponents \(\frac{a^{m}}{a^{n}}=a^{m - n}\) for the \(b\)-terms: \(\frac{b^{2}}{b^{4}}=b^{2-4}=b^{-2}\). So the expression inside the parentheses becomes \(\frac{b^{-2}c^{2}}{3}\).

Step 2: Apply the power of a quotient rule \((\frac{a}{b})^{n}=\frac{a^{n}}{b^{n}}\)

Now, raise the entire fraction to the power of 3: \((\frac{b^{-2}c^{2}}{3})^{3}=\frac{(b^{-2})^{3}(c^{2})^{3}}{3^{3}}\).

Step 3: Apply the power of a power rule \((a^{m})^{n}=a^{mn}\)

For the \(b\)-term: \((b^{-2})^{3}=b^{-2\times3}=b^{-6}\). For the \(c\)-term: \((c^{2})^{3}=c^{2\times3}=c^{6}\). And \(3^{3} = 27\). So now we have \(\frac{b^{-6}c^{6}}{27}\).

Step 4: Rewrite negative exponents as positive exponents in the denominator

Recall that \(a^{-n}=\frac{1}{a^{n}}\), so \(b^{-6}=\frac{1}{b^{6}}\). Thus, \(\frac{b^{-6}c^{6}}{27}=\frac{c^{6}}{27b^{6}}\).

Answer:

\(\frac{c^{6}}{27b^{6}}\) (In the given blanks: the exponent of \(c\) is \(6\), the coefficient is \(27\), and the exponent of \(b\) is \(6\))