A number $$ x $$ varies direatly as $$ y $$ and partly inversly as $$ y^{2} $$ given that $$ x=8 $$, when $$ y=4 $$, and $$ x=25 \cdot 16 $$, when $$ y=5 $$, fend $$ x $$, when $$ y=8 $$.

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To solve the problem, we need to express the relationship between $$ x $$ and $$ y $$ based on the given conditions. The statement "a number $$ x $$ varies directly as $$ y $$ and partly inversely as $$ y^{2} $$" can be expressed mathematically as:

$$
x = k \cdot \frac{y}{y^2} = \frac{k}{y}
$$

where $$ k $$ is a constant.

### Step 1: Find the constant $$ k $$

We have two conditions to find $$ k $$:

1. When $$ y = 4 $$, $$ x = 8 $$:
   $$
   8 = \frac{k}{4} \implies k = 8 \cdot 4 = 32
   $$

2. When $$ y = 5 $$, $$ x = 25 \cdot 16 = 400 $$:
   $$
   400 = \frac{k}{5} \implies k = 400 \cdot 5 = 2000
   $$

### Step 2: Check for consistency

We have two different values for $$ k $$ from the two conditions. Let's check if they are consistent:

1. From the first condition, $$ k = 32 $$.
2. From the second condition, $$ k = 2000 $$.

Since these values are not equal, we need to re-evaluate the relationship.

### Step 3: Re-evaluate the relationship

The relationship should be:

$$
x = k_1 y - \frac{k_2}{y^2}
$$

where $$ k_1 $$ and $$ k_2 $$ are constants.

### Step 4: Set up equations

Using the first condition $$ (x = 8, y = 4) $$:

$$
8 = k_1 \cdot 4 - \frac{k_2}{4^2} \implies 8 = 4k_1 - \frac{k_2}{16} \tag{1}
$$

Using the second condition $$ (x = 400, y = 5) $$:

$$
400 = k_1 \cdot 5 - \frac{k_2}{5^2} \implies 400 = 5k_1 - \frac{k_2}{25} \tag{2}
$$

### Step 5: Solve the system of equations

Now we will solve equations (1) and (2) simultaneously to find $$ k_1 $$ and $$ k_2 $$.

Let's express $$ k_2 $$ from equation (1):

$$
k_2 = 64 - 64k_1 \tag{3}
$$

Substituting equation (3) into equation (2):

$$
400 = 5k_1 - \frac{64 - 64k_1}{25}
$$

Now, let's solve this equation for $$ k_1 $$ and $$ k_2 $$.

### Step 6: Solve for $$ k_1 $$ and $$ k_2 $$

We will solve the equations now.

$$
400 = 5k_1 - \frac{64 - 64k_1}{25}
$$

Multiplying through by 25 to eliminate the fraction:

$$
10000 = 125k_1 - (64 - 64k_1)
$$

This simplifies to:

$$
10000 = 125k_1 - 64 + 64k_1
$$

Combining like terms:

$$
10000 + 64 = 189k_1
$$

$$
10064 = 189k_1
$$

Now, solving for $$ k_1 $$:

$$
k_1 = \frac{10064}{189}
$$

Now substituting $$ k_1 $$ back into equation (3) to find $$ k_2 $$:

$$
k_2 = 64 - 64 \cdot \frac{10064}{189}
$$

Now, let's calculate these values.

### Step 7: Calculate $$ k_1 $$ and $$ k_2 $$

I will calculate $$ k_1 $$ and $$ k_2 $$ now.

$$
k_1 = \frac{10064}{189}
$$

Calculating $$ k_1 $$:

$$
k_1 \approx 53.2
$$

Now substituting back to find $$ k_2 $$:

$$
k_2 = 64 - 64 \cdot 53.2
$$

Calculating $$ k_2 $$:

$$
k_2 \approx -3392
$$

### Step 8: Find $$ x $$ when $$ y = 8 $$

Now we can find $$ x $$ when $$ y = 8 $$:

$$
x = k_1 \cdot 8 - \frac{k_2}{8^2}
$$

Substituting the values of $$ k_1 $$ and $$ k_2 $$:

$$
x = 53.2 \cdot 8 - \frac{-3392}{64}
$$

Calculating $$ x $$:

$$
x = 425.6 + 53.2 = 478.8
$$

Thus, the value of $$ x $$ when $$ y = 8 $$ is approximately $$ 478.8 $$.
When $$ y = 8 $$, $$ x $$ is approximately 478.8.

A number \( x \) varies direatly as \( y \) and partly inversly as \( y^{2} \) given that \( x=8 \), when \( y=4 \), and \( x=25 \cdot 16 \), when \( y=5 \), fend \( x \), when \( y=8 \).

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