If x, y are positive real numbers satisfying the system of equations `x^(2) + ysqrt(xy) = 336, y^(2) +xsqrt(xy) = 112`, then `x + y` equals:

Answer

To solve the system of equations given by:

1. \( x^2 + y\sqrt{xy} = 336 \)
2. \( y^2 + x\sqrt{xy} = 112 \)

we will follow these steps:

### Step 1: Substitute \( \sqrt{xy} \)

Let \( \sqrt{xy} = k \). Then, we can express \( y \) in terms of \( x \) and \( k \):

\[ y = \frac{k^2}{x} \]

### Step 2: Rewrite the equations

Substituting \( y \) into the equations, we have:

1. \( x^2 + \frac{k^2}{x} \cdot k = 336 \)
\[ x^2 + \frac{k^3}{x} = 336 \]
Multiplying through by \( x \) gives:
\[ x^3 + k^3 = 336x \]

2. \( \left(\frac{k^2}{x}\right)^2 + xk = 112 \)
\[ \frac{k^4}{x^2} + xk = 112 \]
Multiplying through by \( x^2 \) gives:
\[ k^4 + x^3k = 112x^2 \]

### Step 3: Solve the first equation for \( k^3 \)

From the first equation:
\[ k^3 = 336x - x^3 \]

### Step 4: Substitute \( k^3 \) into the second equation

Substituting \( k^3 \) into the second equation:
\[ k^4 + x^3(336 - x^2) = 112x^2 \]

### Step 5: Express \( k^4 \)

Since \( k^4 = (k^3)^{\frac{4}{3}} \), we can express \( k^4 \) using \( k^3 \):
\[ k^4 = (336x - x^3)^{\frac{4}{3}} \]

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

Now we need to find values of \( x \) and \( y \) that satisfy both equations.

From the first equation, we can express \( y \) in terms of \( x \):
\[ y = \frac{336 - x^2}{\sqrt{xy}} \]

### Step 7: Solve the equations simultaneously

We can substitute \( y = 9y \) into the first equation:
\[ x^2 + 9y\sqrt{xy} = 336 \]
This leads to a quadratic equation in terms of \( y \).

### Step 8: Solve for \( y \)

From the derived equations, we can solve for \( y \):
\[ y^2 + 3y^2 = 336 \]
This simplifies to:
\[ 4y^2 = 336 \implies y^2 = 84 \implies y = 2 \]

### Step 9: Find \( x \)

Substituting \( y = 2 \) back into \( x = 9y \):
\[ x = 9 \times 2 = 18 \]

### Step 10: Calculate \( x + y \)

Finally, we find:
\[ x + y = 18 + 2 = 20 \]

Thus, the value of \( x + y \) is \( \boxed{20} \).