`x"@"y=(x+y)^(2)`
`x#y=(x-y)^(2)`
Find the value of `(9#7)"@"(21#5):`

Options

A. 67600
B. 62500
C. 57600
D. none of these

Answer

To solve the problem, we need to evaluate the expression \( (9 \# 7) \, @ \, (21 \# 5) \) using the definitions given for the operations \( \# \) and \( @ \).

### Step 1: Calculate \( 9 \# 7 \)
Using the definition of \( x \# y = (x - y)^2 \):
\[ 9 \# 7 = (9 - 7)^2 = 2^2 = 4 \]

### Step 2: Calculate \( 21 \# 5 \)
Using the same definition:
\[ 21 \# 5 = (21 - 5)^2 = 16^2 = 256 \]

### Step 3: Calculate \( 4 \, @ \, 256 \)
Now we need to use the definition of \( x @ y = (x + y)^2 \):
\[ 4 \, @ \, 256 = (4 + 256)^2 = 260^2 \]

### Step 4: Calculate \( 260^2 \)
To find \( 260^2 \):
\[ 260^2 = (26 \times 10)^2 = 26^2 \times 10^2 = 676 \times 100 = 67600 \]

### Final Answer
Thus, the value of \( (9 \# 7) \, @ \, (21 \# 5) \) is:
\[ \boxed{67600} \]