If the sum of squares of two real numbers is 41 and their sum is 9, then the sum of cubes of these two numbers is

Options

A. 169
B. 209
C. 189
D. 198

Answer

To solve the problem, we need to find the sum of cubes of two real numbers \( x \) and \( y \) given that:

1. The sum of their squares is \( x^2 + y^2 = 41 \)
2. The sum of the numbers is \( x + y = 9 \)

We can use the identity for the sum of cubes:

\[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) \]

### Step 1: Calculate \( xy \)

We can find \( xy \) using the square of the sum of the numbers:

\[ (x + y)^2 = x^2 + 2xy + y^2 \]

Substituting the known values:

\[ 9^2 = 41 + 2xy \]

This simplifies to:

\[ 81 = 41 + 2xy \]

Now, subtract 41 from both sides:

\[ 81 - 41 = 2xy \]

\[ 40 = 2xy \]

Dividing both sides by 2 gives:

\[ xy = 20 \]

### Step 2: Substitute values into the sum of cubes formula

Now we have \( x + y = 9 \), \( x^2 + y^2 = 41 \), and \( xy = 20 \). We can now calculate \( x^2 - xy + y^2 \):

\[ x^2 - xy + y^2 = (x^2 + y^2) - xy \]

Substituting the known values:

\[ x^2 - xy + y^2 = 41 - 20 = 21 \]

### Step 3: Calculate \( x^3 + y^3 \)

Now we can substitute back into the sum of cubes formula:

\[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) \]

Substituting the values we have:

\[ x^3 + y^3 = 9 \times 21 \]

Calculating this gives:

\[ x^3 + y^3 = 189 \]

### Final Answer

Thus, the sum of the cubes of the two numbers is:

\[ \boxed{189} \]