The problem states that $a$ varies directly as the square of $b$ and inversely as $c$. We are given that $a=2$ when $b=4$ and $c=24$. We need to find (a) the value of $k$, (b) the value of $a$ when $b=9$ and $c=27$, and (c) the value of $b$ when $a=8$ and $c=6$.

1. Problem Description

The problem states that $a$ varies directly as the square of $b$ and inversely as $c$. We are given that $a=2$ when $b=4$ and $c=24$. We need to find (a) the value of $k$, (b) the value of $a$ when $b=9$ and $c=27$, and (c) the value of $b$ when $a=8$ and $c=6$.

2. Solution Steps

(a) Find the value of $k$.

The relationship between $a$, $b$, and $c$ is given by:

$a = \frac{kb^2}{c}$

Substituting the given values $a=2$, $b=4$, and $c=24$ into the equation:

$2 = \frac{k(4^2)}{24}$

$2 = \frac{16k}{24}$

$2 = \frac{2k}{3}$

Multiplying both sides by 3:

$6 = 2k$

Dividing both sides by 2:

$k = 3$

(b) Find the value of $a$ when $b=9$ and $c=27$.

We know that $k=3$. Substituting $k=3$, $b=9$, and $c=27$ into the equation $a = \frac{kb^2}{c}$:

$a = \frac{3(9^2)}{27}$

$a = \frac{3(81)}{27}$

$a = \frac{243}{27}$

$a = 9$

(c) Find the value of $b$ when $a=8$ and $c=6$.

We know that $k=3$. Substituting $a=8$, $c=6$, and $k=3$ into the equation $a = \frac{kb^2}{c}$:

$8 = \frac{3b^2}{6}$

$8 = \frac{b^2}{2}$

Multiplying both sides by 2:

$16 = b^2$

Taking the square root of both sides:

$b = \pm 4$

Since the problem doesn't specify any restrictions on $b$, we consider both positive and negative values. However, usually, in these types of problems, we consider only the positive root. Therefore, $b=4$.

3. Final Answer

(a) $k=3$

(b) $a=9$

(c) $b=4$