Question

which expression is equivalent to \\(\left(\frac{(3xy^{-5})^{3}}{(x^{-2}y^{2})^{-4}}\
ight)^{-2}\\)? assume \\(x \
eq 0, y \
eq 0\\). \\(\frac{x^{10}y^{14}}{729}\\) \\(\frac{x^{22}}{18y^{46}}\\) \\(\frac{729}{x^{10}y^{14}}\\) \\(\frac{18y^{46}}{x^{22}}\\)

Explanation:

Step1: Simplify the numerator inside the big parentheses

First, simplify \((3xy^{-5})^{3}\). Using the power of a product rule \((ab)^n = a^n b^n\) and the power of a power rule \((a^m)^n = a^{mn}\), we get:
\(3^3 x^3 (y^{-5})^3 = 27x^3 y^{-15}\)

Step2: Simplify the denominator inside the big parentheses

Next, simplify \((x^{-2}y^2)^{-4}\). Using the power of a product rule and the power of a power rule:
\((x^{-2})^{-4}(y^2)^{-4} = x^{8} y^{-8}\)

Step3: Simplify the fraction inside the big parentheses

Now, divide the numerator by the denominator: \(\frac{27x^3 y^{-15}}{x^{8} y^{-8}}\). Using the quotient rule for exponents \(\frac{a^m}{a^n} = a^{m - n}\), we have:
\(27x^{3 - 8} y^{-15 - (-8)} = 27x^{-5} y^{-7}\)

Step4: Apply the outer exponent of -2

Now, apply the exponent -2 to the result from Step 3: \((27x^{-5} y^{-7})^{-2}\). Using the power of a product rule and the power of a power rule:
\(27^{-2} x^{(-5)\times(-2)} y^{(-7)\times(-2)} = \frac{1}{27^2} x^{10} y^{14}\)

Step5: Calculate \(27^2\)

Since \(27^2 = 729\), the expression becomes \(\frac{x^{10} y^{14}}{729}\)

Answer:

\(\frac{x^{10}y^{14}}{729}\) (the first option in the top - left box)