The problem consists of two parts. (a) We are given the equation $y = (\frac{pr}{m} - p^2r)^{-\frac{3}{2}}$ and asked to make $r$ the subject of the formula. (b) We are asked to find the value of $r$ when $y = -8$, $m = 1$, and $p = 3$.

1. Problem Description

The problem consists of two parts.

(a) We are given the equation $y = (\frac{pr}{m} - p^2r)^{-\frac{3}{2}}$ and asked to make $r$ the subject of the formula.

(b) We are asked to find the value of $r$ when $y = -8$, $m = 1$, and $p = 3$.

2. Solution Steps

(a) To make $r$ the subject of the formula $y = (\frac{pr}{m} - p^2r)^{-\frac{3}{2}}$, we can proceed as follows:

First, raise both sides to the power of $-\frac{2}{3}$:

$y^{-\frac{2}{3}} = (\frac{pr}{m} - p^2r)^{-\frac{3}{2} \cdot (-\frac{2}{3})}$

$y^{-\frac{2}{3}} = \frac{pr}{m} - p^2r$

Factor out $r$ on the right side:

$y^{-\frac{2}{3}} = r(\frac{p}{m} - p^2)$

Now, divide both sides by $(\frac{p}{m} - p^2)$:

$r = \frac{y^{-\frac{2}{3}}}{\frac{p}{m} - p^2}$

$r = \frac{1}{y^{\frac{2}{3}}(\frac{p}{m} - p^2)}$

$r = \frac{1}{(\frac{p}{m} - p^2)y^{\frac{2}{3}}}$

(b) To find the value of $r$ when $y = -8$, $m = 1$, and $p = 3$, we substitute these values into the equation obtained in part (a):

$r = \frac{1}{(\frac{3}{1} - 3^2)(-8)^{\frac{2}{3}}}$

$r = \frac{1}{(3 - 9)((-8)^{\frac{1}{3}})^2}$

$r = \frac{1}{(-6)(-2)^2}$

$r = \frac{1}{(-6)(4)}$

$r = \frac{1}{-24}$

$r = -\frac{1}{24}$

3. Final Answer

(a) $r = \frac{1}{(\frac{p}{m} - p^2)y^{\frac{2}{3}}}$

(b) $r = -\frac{1}{24}$