The problem gives the equation $y = (\frac{pr}{m} - p^2r)^{-\frac{3}{2}}$. Part (a) asks to make $r$ the subject of the formula. Part (b) asks to find the value of $r$ when $y = -8$, $m = 1$, and $p = 3$.

1. Problem Description

The problem gives the equation $y = (\frac{pr}{m} - p^2r)^{-\frac{3}{2}}$. Part (a) asks to make $r$ the subject of the formula. Part (b) asks to find the value of $r$ when $y = -8$, $m = 1$, and $p = 3$.

2. Solution Steps

(a) Make $r$ the subject.

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

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

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

Factor out $r$ from the right side:

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

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

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

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

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

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

(b) Find the value of $r$ when $y = -8$, $m = 1$, and $p = 3$.

Substitute the given values into the equation $r = \frac{m}{y^{\frac{2}{3}}(p - p^2m)}$:

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

First, $(-8)^{\frac{2}{3}} = ((-8)^{\frac{1}{3}})^2 = (-2)^2 = 4$.

Next, $3 - 3^2(1) = 3 - 9(1) = 3 - 9 = -6$.

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

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

3. Final Answer

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

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