`x"@"y=(x+y)^(2)`
`x#y=(x-y)^(2)`
Find the value of `516#(14"@"13)):`

Options

A. 4539
B. 42336
C. 45369
D. 89719

Answer

To solve the problem, we need to evaluate the expression \( 516 \# (14 \, @ \, 13) \) using the defined operations \( x \, @ \, y = (x + y)^2 \) and \( x \# y = (x - y)^2 \).

### Step 1: Calculate \( 14 \, @ \, 13 \)

Using the operation \( x \, @ \, y = (x + y)^2 \):

1. Substitute \( x = 14 \) and \( y = 13 \):
\[ 14 \, @ \, 13 = (14 + 13)^2 \]

2. Calculate \( 14 + 13 \):
\[ 14 + 13 = 27 \]

3. Now square the result:
\[ (27)^2 = 729 \]

So, \( 14 \, @ \, 13 = 729 \).

### Step 2: Calculate \( 516 \# 729 \)

Now we need to use the operation \( x \# y = (x - y)^2 \):

1. Substitute \( x = 516 \) and \( y = 729 \):
\[ 516 \# 729 = (516 - 729)^2 \]

2. Calculate \( 516 - 729 \):
\[ 516 - 729 = -213 \]

3. Now square the result:
\[ (-213)^2 = 45369 \]

So, \( 516 \# 729 = 45369 \).

### Final Answer

The value of \( 516 \# (14 \, @ \, 13) \) is \( 45369 \).

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