Question

Express the following fraction in simplest form, only using positive exponents.
start fraction, 3, b, to the power minus 9 , divided by, left parenthesis, 3, b, to the power minus 4 , c, to the power minus 3 , right parenthesis, to the power minus 3 , end fraction
(3b
−4
c
−3
)
−3

3b
−9

​

Answer 1

To simplify the given fraction \(\frac{3b^{-9}}{(3b^{-4}c^{-3})^{-3}}\), we will first simplify the denominator.

### Step 1: Simplify the Denominator

The denominator is \((3b^{-4}c^{-3})^{-3}\). To simplify this, we use the property of exponents that states \((x^m)^{-n} = x^{-m \cdot n}\).

So we apply this property:

\[
(3b^{-4}c^{-3})^{-3} = 3^{-3} (b^{-4})^{-3} (c^{-3})^{-3}
\]

Calculating each part:

1. \(3^{-3} = \frac{1}{27}\)
2. \((b^{-4})^{-3} = b^{12}\)
3. \((c^{-3})^{-3} = c^{9}\)

Combining these:

\[
(3b^{-4}c^{-3})^{-3} = \frac{1}{27} b^{12} c^{9}
\]

### Step 2: Rewrite the Denominator

Putting it all together, we have:

\[
(3b^{-4}c^{-3})^{-3} = \frac{b^{12} c^{9}}{27}
\]

### Step 3: Rewrite the Whole Expression

Now we rewrite the entire expression:

\[
\frac{3b^{-9}}{\frac{b^{12} c^{9}}{27}} = 3b^{-9} \cdot \frac{27}{b^{12}c^{9}} = \frac{3 \cdot 27 b^{-9}}{b^{12}c^{9}}
\]

Simplifying it further:

\[
= \frac{81 b^{-9}}{b^{12} c^{9}}
\]

### Step 4: Combine the Exponents of \(b\)

Using the property \(b^{m} / b^{n} = b^{m-n}\):

\[
= \frac{81}{b^{12 - 9} c^{9}} = \frac{81}{b^{3} c^{9}}
\]

### Conclusion

Thus, the fraction in simplest form, using only positive exponents, is:

\[
\frac{81}{b^3 c^9}
\]