Question

which expression is equivalent to \\(\frac{(3m^{-3}n)^{-3}}{6mn^{-2}}\\)? assume \\(m \
eq 0, n \
eq 0\\).
options: \\(\frac{1}{2m^8n}\\), \\(\frac{m^8}{162n}\\), \\(\frac{4m^8}{3n^3}\\), \\(\frac{8m^9}{n^3}\\)

Explanation:

Step1: Apply power of a product rule to the numerator

The numerator is \((3m^{-3}n)^{-3}\). Using the power of a product rule \((ab)^n = a^n b^n\), we get \(3^{-3}(m^{-3})^{-3}n^{-3}\).
Simplify the exponents: \(3^{-3}m^{9}n^{-3}\) (since \((a^m)^n=a^{mn}\), so \((m^{-3})^{-3}=m^{(-3)\times(-3)} = m^9\)).

Step2: Rewrite the original expression with the simplified numerator

The original expression is \(\frac{(3m^{-3}n)^{-3}}{6mn^{-2}}=\frac{3^{-3}m^{9}n^{-3}}{6mn^{-2}}\).

Step3: Separate the coefficients and variables

Separate the coefficients and the variables with \(m\) and \(n\): \(\frac{3^{-3}}{6}\times\frac{m^{9}}{m}\times\frac{n^{-3}}{n^{-2}}\).

Step4: Simplify the coefficients

Simplify \(\frac{3^{-3}}{6}\). We know that \(3^{-3}=\frac{1}{3^3}=\frac{1}{27}\), so \(\frac{\frac{1}{27}}{6}=\frac{1}{27\times6}=\frac{1}{162}\)? Wait, no, wait. Wait, maybe I made a mistake here. Wait, let's check again. Wait, the coefficient in the numerator after step 1 is \(3^{-3}\) and the denominator coefficient is 6. So \(\frac{3^{-3}}{6}=\frac{1}{3^3\times6}=\frac{1}{27\times6}=\frac{1}{162}\)? But that doesn't match the options. Wait, maybe I messed up the original problem. Wait, the original problem is \(\frac{(3m^{-3}n)^{-3}}{6mn^{-2}}\)? Wait, maybe the numerator is \((3m^{-3}n)^{-3}\) and the denominator is \(6mn^{-2}\). Wait, let's re - do the coefficient part. Wait, \(3^{-3}=\frac{1}{27}\), and the denominator coefficient is 6. So \(\frac{3^{-3}}{6}=\frac{1}{27\times6}=\frac{1}{162}\)? But the options have \(\frac{m^8}{162n}\)? Wait, no, maybe I made a mistake in the variable part. Wait, let's do the variable part first.

For the \(m\) terms: \(\frac{m^{9}}{m}=m^{9 - 1}=m^{8}\) (using the rule \(\frac{a^m}{a^n}=a^{m - n}\)).

For the \(n\) terms: \(\frac{n^{-3}}{n^{-2}}=n^{-3-(-2)}=n^{-1}\) (using the rule \(\frac{a^m}{a^n}=a^{m - n}\)).

Now the coefficient: \(\frac{3^{-3}}{6}=\frac{1}{27\times6}=\frac{1}{162}\).

So putting it all together: \(\frac{1}{162}\times m^{8}\times n^{-1}=\frac{m^{8}}{162n}\) (since \(n^{-1}=\frac{1}{n}\)).

Wait, but let's check the options. One of the options is \(\frac{m^{8}}{162n}\), which is the second option. Wait, maybe I made a mistake in the initial step. Let's re - derive:

1. Expand \((3m^{-3}n)^{-3}\):

• Using \((abc)^k=a^k b^k c^k\), we have \(3^{-3}(m^{-3})^{-3}n^{-3}\).
• \((m^{-3})^{-3}=m^{(-3)\times(-3)} = m^9\), so \((3m^{-3}n)^{-3}=3^{-3}m^{9}n^{-3}=\frac{1}{27}m^{9}n^{-3}\).

1. Now the expression is \(\frac{\frac{1}{27}m^{9}n^{-3}}{6mn^{-2}}\).
2. Divide the coefficients: \(\frac{\frac{1}{27}}{6}=\frac{1}{27\times6}=\frac{1}{162}\).
3. Divide the \(m\) terms: \(\frac{m^{9}}{m}=m^{9 - 1}=m^{8}\) (using \(\frac{a^m}{a^n}=a^{m - n}\)).
4. Divide the \(n\) terms: \(\frac{n^{-3}}{n^{-2}}=n^{-3-(-2)}=n^{-1}=\frac{1}{n}\) (using \(\frac{a^m}{a^n}=a^{m - n}\)).

Multiply all the parts together: \(\frac{1}{162}\times m^{8}\times\frac{1}{n}=\frac{m^{8}}{162n}\).

Answer:

B. \(\frac{m^{8}}{162n}\) (assuming the second option is labeled B)