Given that `(a^(2)+b^(2))=60`, then find the value of `(a+b)^(2)+ (a-b)^(2)`
A 90 B 120 C 140 D 150

Options

A. C
B. A
C. D
D. B

Answer

To solve the problem, we need to find the value of \((a+b)^2 + (a-b)^2\) given that \(a^2 + b^2 = 60\).

### Step-by-Step Solution:

1. **Use the identities for squares**:
We know the following identities:
- \((a+b)^2 = a^2 + 2ab + b^2\)
- \((a-b)^2 = a^2 - 2ab + b^2\)

2. **Add the two identities**:
Now, we can add these two equations:
\[ (a+b)^2 + (a-b)^2 = (a^2 + 2ab + b^2) + (a^2 - 2ab + b^2) \]

3. **Simplify the expression**:
When we combine the terms, we get:
\[ (a+b)^2 + (a-b)^2 = a^2 + 2ab + b^2 + a^2 - 2ab + b^2 \]
The \(2ab\) and \(-2ab\) terms cancel each other out:
\[ = a^2 + b^2 + a^2 + b^2 = 2a^2 + 2b^2 \]

4. **Factor out the common term**:
We can factor out the 2:
\[ = 2(a^2 + b^2) \]

5. **Substitute the given value**:
We know from the problem that \(a^2 + b^2 = 60\). Therefore:
\[ = 2 \times 60 = 120 \]

### Final Answer:
Thus, the value of \((a+b)^2 + (a-b)^2\) is \(120\).

### Answer Options:
- A) 90
- B) 120
- C) 140
- D) 150

The correct answer is **B) 120**.