Question

express the following fraction in simplest form, only using positive exponents.
\frac{(3d^{3}z)^{-3}}{5d^{-3}z^{4}}

Explanation:

Step1: Apply the negative exponent rule

The negative exponent rule states that \(a^{-n}=\frac{1}{a^{n}}\) and \(\frac{1}{a^{-n}} = a^{n}\). So, we can rewrite the numerator \((3d^{3}z)^{-3}\) as \(\frac{1}{(3d^{3}z)^{3}}\) and the denominator has \(d^{-3}\) which can be rewritten as \(\frac{1}{d^{3}}\) in the numerator. So the expression becomes:

$$ \frac{\frac{1}{(3d^{3}z)^{3}}}{5\times\frac{1}{d^{3}}z^{4}}=\frac{1}{(3d^{3}z)^{3}}\times\frac{d^{3}}{5z^{4}} $$

Step2: Expand the power in the denominator

Expand \((3d^{3}z)^{3}\) using the power of a product rule \((ab)^{n}=a^{n}b^{n}\). So \((3d^{3}z)^{3}=3^{3}(d^{3})^{3}z^{3}=27d^{9}z^{3}\). Now our expression is:

$$ \frac{d^{3}}{27d^{9}z^{3}\times5z^{4}} $$

Step3: Simplify the exponents of like bases

For the \(d\) terms, use the quotient rule of exponents \(a^{m}\div a^{n}=a^{m - n}\). So \(d^{3}\div d^{9}=d^{3-9}=d^{-6}\). For the \(z\) terms, \(z^{3}\times z^{4}=z^{3 + 4}=z^{7}\) (using the product rule of exponents \(a^{m}\times a^{n}=a^{m + n}\)). And the constants are \(27\times5 = 135\). So now we have:

$$ \frac{d^{-6}}{135z^{7}} $$

Step4: Convert negative exponents to positive

Using the negative exponent rule again, \(d^{-6}=\frac{1}{d^{6}}\). So the expression becomes:

$$ \frac{1}{135d^{6}z^{7}} $$

Answer:

\(\frac{1}{135d^{6}z^{7}}\)