`(x+y)^(2//3) +2(x-y)^(2//3) =3(x^(2)-y^(2))^(1//3), 3x-2y =13`

Options

A. `((13)/(3),0)`
B. (9,7)
C. `(0,-(13)/(2))`
D. none

Answer

To solve the given equations:

1. \((x+y)^{\frac{2}{3}} + 2(x-y)^{\frac{2}{3}} = 3(x^2 - y^2)^{\frac{1}{3}}\)
2. \(3x - 2y = 13\)

We will follow these steps:

### Step 1: Rewrite the first equation
We can rewrite the first equation using the identity \(x^2 - y^2 = (x+y)(x-y)\):
\[ (x+y)^{\frac{2}{3}} + 2(x-y)^{\frac{2}{3}} = 3((x+y)(x-y))^{\frac{1}{3}} \]

### Step 2: Introduce a substitution
Let:
\[ t = \frac{x+y}{x-y} \]
Then we can express \(x+y\) and \(x-y\) in terms of \(t\):
\[ x+y = t(x-y) \]

### Step 3: Substitute into the equation
Substituting \(x+y\) and \(x-y\) into the equation gives:
\[ (t(x-y))^{\frac{2}{3}} + 2(x-y)^{\frac{2}{3}} = 3((t(x-y))(x-y))^{\frac{1}{3}} \]
This simplifies to:
\[ t^{\frac{2}{3}}(x-y)^{\frac{2}{3}} + 2(x-y)^{\frac{2}{3}} = 3(t(x-y)^{2})^{\frac{1}{3}} \]

### Step 4: Factor out \((x-y)^{\frac{2}{3}}\)
Factoring out \((x-y)^{\frac{2}{3}}\):
\[ (x-y)^{\frac{2}{3}}(t^{\frac{2}{3}} + 2) = 3(t^{\frac{1}{3}}(x-y)) \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ (x-y)^{\frac{2}{3}}(t^{\frac{2}{3}} + 2) - 3t^{\frac{1}{3}}(x-y) = 0 \]

### Step 6: Solve for \(x-y\)
This can be solved by setting \((x-y)^{\frac{2}{3}} = 0\) or solving the quadratic in \(t\):
1. If \((x-y)^{\frac{2}{3}} = 0\), then \(x = y\).
2. Otherwise, we can solve for \(t\).

### Step 7: Solve the second equation
Using the second equation \(3x - 2y = 13\), we can substitute \(y = 0\) (from \(x = y\)):
\[ 3x - 0 = 13 \implies x = \frac{13}{3}, y = 0 \]

### Step 8: Solve for the other case
Now consider \(t = 2\):
\[ \frac{x+y}{x-y} = 2 \implies x+y = 2(x-y) \]
This gives:
\[ x+y = 2x - 2y \implies 3y = x \implies x = 3y \]

### Step 9: Substitute into the second equation
Substituting \(x = 3y\) into \(3x - 2y = 13\):
\[ 3(3y) - 2y = 13 \implies 9y - 2y = 13 \implies 7y = 13 \implies y = \frac{13}{7} \]
Then substituting back to find \(x\):
\[ x = 3y = 3 \cdot \frac{13}{7} = \frac{39}{7} \]

### Final Solutions
The solutions are:
1. \((x, y) = \left(\frac{13}{3}, 0\right)\)
2. \((x, y) = \left(\frac{39}{7}, \frac{13}{7}\right)\)