Two positive numbers x and y are such that `x gt y`. If the difference of these numbers is 5 and their product is 24 find :
(i) sum of these number
(ii) difference of their cubes
(iii) Sum of their cubes

Answer

To solve the problem, we need to find two positive numbers \( x \) and \( y \) such that \( x > y \), the difference between the numbers is 5, and their product is 24. We will find the following:

1. The sum of these numbers.
2. The difference of their cubes.
3. The sum of their cubes.

Let's break this down step by step.

### Step 1: Set up the equations

From the problem, we have the following two equations:

1. \( x - y = 5 \) (Equation 1)
2. \( xy = 24 \) (Equation 2)

### Step 2: Express \( x \) in terms of \( y \)

From Equation 1, we can express \( x \) in terms of \( y \):

\[ x = y + 5 \]

### Step 3: Substitute \( x \) in Equation 2

Now, we substitute \( x \) in Equation 2:

\[ (y + 5)y = 24 \]

Expanding this gives:

\[ y^2 + 5y = 24 \]

### Step 4: Rearrange the equation

Rearranging the equation gives us a standard quadratic form:

\[ y^2 + 5y - 24 = 0 \]

### Step 5: Factor the quadratic equation

Now, we need to factor the quadratic equation. We look for two numbers that multiply to \(-24\) and add to \(5\). The numbers \(8\) and \(-3\) satisfy this:

\[ (y + 8)(y - 3) = 0 \]

### Step 6: Solve for \( y \)

Setting each factor to zero gives us:

\[ y + 8 = 0 \quad \text{or} \quad y - 3 = 0 \]

Thus,

\[ y = -8 \quad \text{(not valid since } y \text{ must be positive)} \quad \text{or} \quad y = 3 \]

### Step 7: Find \( x \)

Now that we have \( y = 3 \), we can find \( x \):

\[ x = y + 5 = 3 + 5 = 8 \]

### Step 8: Calculate the sum of the numbers

Now we can find the sum of the numbers:

\[ x + y = 8 + 3 = 11 \]

### Step 9: Calculate the difference of their cubes

Using the identity \( x^3 - y^3 = (x - y)(x^2 + xy + y^2) \):

1. \( x - y = 5 \)
2. \( x^2 = 8^2 = 64 \)
3. \( y^2 = 3^2 = 9 \)
4. \( xy = 24 \)

Now calculate:

\[ x^2 + xy + y^2 = 64 + 24 + 9 = 97 \]

So,

\[ x^3 - y^3 = 5 \times 97 = 485 \]

### Step 10: Calculate the sum of their cubes

Using the identity \( x^3 + y^3 = (x + y)(x^2 - xy + y^2) \):

1. \( x + y = 11 \)
2. \( x^2 + y^2 = 64 + 9 = 73 \)

Now calculate:

\[ x^2 - xy + y^2 = 73 - 24 = 49 \]

So,

\[ x^3 + y^3 = 11 \times 49 = 539 \]

### Final Answers

1. The sum of the numbers \( x + y = 11 \)
2. The difference of their cubes \( x^3 - y^3 = 485 \)
3. The sum of their cubes \( x^3 + y^3 = 539 \)