describe how you would simplify the given expression.\n$left(\frac{20x^{5}y^{2}}{5x^{-3}y^{7}}\right)^{-3},x\…

Explanation:

Step1: Simplify the fraction inside the parentheses

First, divide the coefficients and use the quotient - rule of exponents $a^m\div a^n=a^{m - n}$. The coefficient $20\div5 = 4$. For the $x$ - terms, $x^{5}\div x^{-3}=x^{5-(-3)}=x^{8}$. For the $y$ - terms, $y^{2}\div y^{7}=y^{2 - 7}=y^{-5}$. So the fraction inside the parentheses is $4x^{8}y^{-5}$.

Step2: Apply the power - of - a - power rule

Use the rule $(ab)^n=a^n b^n$ and $(a^m)^n=a^{mn}$. So $(4x^{8}y^{-5})^{-3}=4^{-3}x^{8\times(-3)}y^{-5\times(-3)}$.

Step3: Calculate the powers

$4^{-3}=\frac{1}{4^{3}}=\frac{1}{64}$, $x^{8\times(-3)}=x^{-24}$, and $y^{-5\times(-3)}=y^{15}$.

Step4: Rewrite with positive exponents

Using the rule $a^{-n}=\frac{1}{a^{n}}$, we get $\frac{y^{15}}{64x^{24}}$.

Answer:

$\frac{y^{15}}{64x^{24}}$