If `2x-3y=10` and `xy=16`, find the value of `8x^(3)-27y^(3)`.

Answer

To find the value of \(8x^3 - 27y^3\) given the equations \(2x - 3y = 10\) and \(xy = 16\), we can follow these steps:

### Step 1: Cube the first equation
We start by cubing the equation \(2x - 3y = 10\):
\[ (2x - 3y)^3 = 10^3 \]
This gives us:
\[ (2x - 3y)^3 = 1000 \]

### Step 2: Apply the formula for the cube of a binomial
Using the formula \((a - b)^3 = a^3 - b^3 - 3ab(a - b)\), where \(a = 2x\) and \(b = 3y\), we can expand:
\[ (2x - 3y)^3 = (2x)^3 - (3y)^3 - 3(2x)(3y)(2x - 3y) \]
This simplifies to:
\[ 8x^3 - 27y^3 - 18xy(2x - 3y) = 1000 \]

### Step 3: Substitute known values
We know \(xy = 16\) and \(2x - 3y = 10\). Substitute these values into the equation:
\[ 8x^3 - 27y^3 - 18(16)(10) = 1000 \]

### Step 4: Calculate the product
Calculate \(18 \times 16 \times 10\):
\[ 18 \times 16 = 288 \quad \text{and} \quad 288 \times 10 = 2880 \]
So the equation becomes:
\[ 8x^3 - 27y^3 - 2880 = 1000 \]

### Step 5: Isolate \(8x^3 - 27y^3\)
Now, we can isolate \(8x^3 - 27y^3\):
\[ 8x^3 - 27y^3 = 1000 + 2880 \]
Calculating the right side:
\[ 1000 + 2880 = 3880 \]

### Final Answer
Thus, the value of \(8x^3 - 27y^3\) is:
\[ \boxed{3880} \]